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Question

(A) The electric potential $V$ due to an infinitely long line charge with linear charge density $\lambda$ at a distance $r$ is given by $V = -2k\lambd

Vedclass · Accepted Answer

$\frac{r_1 r_2}{r_3^2} = 	ext{constant}$

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(A) The electric potential $V$ due to an infinitely long line charge with linear charge density $\lambda$ at a distance $r$ is given by $V = -2k\lambda \ln(r) + C$,where $k = \frac{1}{4\pi\epsilon_0}$.For three line charges with densities $\lambda_1 = \lambda$,$\lambda_2 = \lambda$,and $\lambda_3 = -2\lambda$,the total potential $V$ at a point is the sum of individual potentials:$V = -2k\lambda \ln(r_1) - 2k\lambda \ln(r_2) - 2k(-2\lambda) \ln(r_3) + C'$$V = -2k\lambda [\ln(r_1) + \ln(r_2) - 2\ln(r_3)] + C'$$V = -2k\lambda [\ln(r_1 r_2) - \ln(r_3^2)] + C'$$V = -2k\lambda \ln\left(\frac{r_1 r_2}{r_3^2}ight) + C'$For equipotential points,$V$ must be constant. Therefore,$\ln\left(\frac{r_1 r_2}{r_3^2}ight)$ must be constant,which implies $\frac{r_1 r_2}{r_3^2} = 	ext{constant}$.

Three infinitely long linear charges of charge density $\lambda$,$\lambda$,and $-2\lambda$ are placed in space. $A$ point in space is specified by its perpendicular distances $r_1$,$r_2$,and $r_3$ respectively from the linear charges. For the points which are equipotential:

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