Mixing
The derivative of the Electric field for a uniformly charged rod
$begingroup$
The formula for the electric field at a point due to a charge $Q$ (just considering the magnitude) at some distance $x$ away from the point is $E=dfrac{k_eQ}{x^2}$ where $k_e$ is a constant equal to approximately $8.99 times 10^{9}$.
If we now consider a uniformly charged rod with charge density $lambda$ and length $L$, then the charge of the entire rod is $lambda L$.
Assuming a point $P$ that is $a$ units from the end of the rod along the $x$-axis, how would I express the derivative of the electric field with respect to $x$ when the charge $Q$ is a function of $x$, but the electric field $E$ varies with both $Q$ and the inverse square of the distance between $x$ and the point $P$? Isn't $E$ a function of two variables in this case?
derivatives
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
$endgroup$
add a comment |
$begingroup$
The formula for the electric field at a point due to a charge $Q$ (just considering the magnitude) at some distance $x$ away from the point is $E=dfrac{k_eQ}{x^2}$ where $k_e$ is a constant equal to approximately $8.99 times 10^{9}$.
If we now consider a uniformly charged rod with charge density $lambda$ and length $L$, then the charge of the entire rod is $lambda L$.
Assuming a point $P$ that is $a$ units from the end of the rod along the $x$-axis, how would I express the derivative of the electric field with respect to $x$ when the charge $Q$ is a function of $x$, but the electric field $E$ varies with both $Q$ and the inverse square of the distance between $x$ and the point $P$? Isn't $E$ a function of two variables in this case?
derivatives
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
$endgroup$
add a comment |
$begingroup$
The formula for the electric field at a point due to a charge $Q$ (just considering the magnitude) at some distance $x$ away from the point is $E=dfrac{k_eQ}{x^2}$ where $k_e$ is a constant equal to approximately $8.99 times 10^{9}$.
If we now consider a uniformly charged rod with charge density $lambda$ and length $L$, then the charge of the entire rod is $lambda L$.
Assuming a point $P$ that is $a$ units from the end of the rod along the $x$-axis, how would I express the derivative of the electric field with respect to $x$ when the charge $Q$ is a function of $x$, but the electric field $E$ varies with both $Q$ and the inverse square of the distance between $x$ and the point $P$? Isn't $E$ a function of two variables in this case?
derivatives
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
$endgroup$
The formula for the electric field at a point due to a charge $Q$ (just considering the magnitude) at some distance $x$ away from the point is $E=dfrac{k_eQ}{x^2}$ where $k_e$ is a constant equal to approximately $8.99 times 10^{9}$.
If we now consider a uniformly charged rod with charge density $lambda$ and length $L$, then the charge of the entire rod is $lambda L$.
Assuming a point $P$ that is $a$ units from the end of the rod along the $x$-axis, how would I express the derivative of the electric field with respect to $x$ when the charge $Q$ is a function of $x$, but the electric field $E$ varies with both $Q$ and the inverse square of the distance between $x$ and the point $P$? Isn't $E$ a function of two variables in this case?
derivatives
derivatives
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
edited Sep 1 '16 at 7:58
Ng Chung Tak
14.7k31334
14.7k31334
asked Sep 1 '16 at 2:06
juliebjulieb
12
asked Sep 1 '16 at 2:06
juliebjulieb
12
asked Sep 1 '16 at 2:06
juliebjulieb
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's write $Q$ as a function of distance $R$ from the particle:
$$Q=int_x^R lambda dR$$
This tells us that
$$frac{dQ}{dR}=lambdato dQ=lambda dRtag{1}$$
Now,
$$E=k_efrac{Q}{R^2}to E=int_0^EdE=int_0^Qk_efrac{dQ}{R^2}tag{2}$$
Substitution of $(1)$ into $(2)$ yields
$$E=k_eint_x^{x+L}frac{lambda}{R^2}dR$$
Can you go from here?
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
$endgroup$
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let's write $Q$ as a function of distance $R$ from the particle:
$$Q=int_x^R lambda dR$$
This tells us that
$$frac{dQ}{dR}=lambdato dQ=lambda dRtag{1}$$
Now,
$$E=k_efrac{Q}{R^2}to E=int_0^EdE=int_0^Qk_efrac{dQ}{R^2}tag{2}$$
Substitution of $(1)$ into $(2)$ yields
$$E=k_eint_x^{x+L}frac{lambda}{R^2}dR$$
Can you go from here?
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
$endgroup$
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
add a comment |
$begingroup$
Let's write $Q$ as a function of distance $R$ from the particle:
$$Q=int_x^R lambda dR$$
This tells us that
$$frac{dQ}{dR}=lambdato dQ=lambda dRtag{1}$$
Now,
$$E=k_efrac{Q}{R^2}to E=int_0^EdE=int_0^Qk_efrac{dQ}{R^2}tag{2}$$
Substitution of $(1)$ into $(2)$ yields
$$E=k_eint_x^{x+L}frac{lambda}{R^2}dR$$
Can you go from here?
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
$endgroup$
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
add a comment |
$begingroup$
Let's write $Q$ as a function of distance $R$ from the particle:
$$Q=int_x^R lambda dR$$
This tells us that
$$frac{dQ}{dR}=lambdato dQ=lambda dRtag{1}$$
Now,
$$E=k_efrac{Q}{R^2}to E=int_0^EdE=int_0^Qk_efrac{dQ}{R^2}tag{2}$$
Substitution of $(1)$ into $(2)$ yields
$$E=k_eint_x^{x+L}frac{lambda}{R^2}dR$$
Can you go from here?
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
$endgroup$
Let's write $Q$ as a function of distance $R$ from the particle:
$$Q=int_x^R lambda dR$$
This tells us that
$$frac{dQ}{dR}=lambdato dQ=lambda dRtag{1}$$
Now,
$$E=k_efrac{Q}{R^2}to E=int_0^EdE=int_0^Qk_efrac{dQ}{R^2}tag{2}$$
Substitution of $(1)$ into $(2)$ yields
$$E=k_eint_x^{x+L}frac{lambda}{R^2}dR$$
Can you go from here?
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
answered Sep 1 '16 at 2:50
HDE 226868HDE 226868
2,1311130
2,1311130
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
add a comment |
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator?
$endgroup$
– julieb
Sep 1 '16 at 21:10
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
@julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience".
$endgroup$
– HDE 226868
Sep 2 '16 at 15:11
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something?
$endgroup$
– julieb
Sep 2 '16 at 19:51
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
I meant $(L + a -x)^2$
$endgroup$
– julieb
Sep 2 '16 at 20:00
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
$begingroup$
@julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=int_a^{a+x}lambda dx$. If $a=L$, then the charge is the total charge on the rod, $lambda L$.
$endgroup$
– HDE 226868
Sep 3 '16 at 1:20
add a comment |
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