Question

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on the x axis, and a 70-nc charge at y = −6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Answer 1

Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

$V = V_1 + V_2 + V_3$

here we have potential due to 50 nC placed at y = 6 m

$V_1 = \frac{kQ}{r}$

$V_1 = \frac{(9\times 10^9)(50 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_1 = 45 Volts$

Now potential due to -80 nC charge placed at x = -4

$V_2 = \frac{kQ}{r}$

$V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}$

$V_2 = -60 Volts$

Now potential due to 70 nC placed at y = -6 m

$V_3 = \frac{kQ}{r}$

$V_3 = \frac{(9\times 10^9)(70 \times 10^{-9})}{\sqrt{6^2 + 8^2}}$

$V_3 = 63 Volts$

Now total potential at this point is given as

$V = 45 - 60 + 63 = 48 Volts$