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eemg
eemg
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I am suspicious of the answer given above by Sam.

Suppose the radius of the shell is $a$. Then, the expression of the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., outside the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ inside the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.

I am suspicious of the answer given above by Sam.

Suppose the radius of the shell is $a$. Then, the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., outside the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ inside the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.

I am suspicious of the answer given above by Sam.

Suppose the radius of the shell is $a$. Then, the expression of the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., outside the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ inside the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.

Source Link
eemg
eemg
  • 50
  • 5

I am suspicious of the answer given above by Sam.

Suppose the radius of the shell is $a$. Then, the contribution of the shell, $V_{shell}=\frac{kx}{r}$, to the total potential is only true for $r>a$, i.e., outside the shell. Now, the problem says that the shell is grounded, and that means that $V=0$ inside the shell. So we cannot simply set $V=\frac{kx}{r}+\frac{kq}{r}$ to zero.