Question

Four point charges, each of charge 2. 5 x 10-5 c, are located on the x- and y-axes, one at each of the locations (0, 2. 0 m), (0, -2. 0 m), (2. 0 m, 0), and (-2. 0 m, 0). the potential at the origin is__________.

Answer 1

The Potential at the origin due to the four point charges located at  (0, 2), (0, -2), (2 , 0), and (-2 , 0) will be - $45\times10^{4}$ volts.

We have four point charges, each of charge $2.5\times10^{-5}$ C, are located on the x - axis and y - axis, one at each of the locations (0, 2), (0, -2), (2 , 0), and (-2 , 0).

We have to calculate the potential at the origin O(0,0).

What is the formula to calculate the Electric potential at a point ' r ' meters away from a point charge ' q ' ?

The Electric field potential due to point charge ' q ' at a distance ' r ' meters apart is -

$V(r) = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}$

Now, in the question given to us -

q = $2.5\times10^{-5}$

Let us label the four points as - A (0, 2), B (0, -2), C (2 , 0), and D (-2 , 0).

Now, the potential due to charge located at point A will be -

$V_{A} = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}\\V_{A} = \frac{1}{4\pi\epsilon_{o} } \frac{q}{r}\\V_{A} = 9\times10^{9}\times\frac{2.5\times10^{-5} }{2} \\V_{A} = 11.25\times10^{4}$

Now, the distance of the center from each point will be same = 2 m.

Hence, the potential due to each charge at the origin will be -

$V_{A} =V_{B} =V_{C} =V_{D} = 11.25\times10^{4}$

Since, the potential is a scalar quantity, we can add the potential due to each charge using simple addition -

$V_{A} +V_{B} +V_{C} +V_{D} = 4\times11.25\times= 45\times10^{4}$ Volts

Hence, the potential at the center will be - $45\times10^{4}$ volts.

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