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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?

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Answer:

V = 48 Volts

Step-by-step 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 = (kQ)/(r)


V_1 = ((9* 10^9)(50 * 10^(-9)))/(√(6^2 + 8^2))


V_1 = 45 Volts

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


V_2 = (kQ)/(r)


V_2 = ((9* 10^9)(-80 * 10^(-9)))/(12)


V_2 = -60 Volts

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


V_3 = (kQ)/(r)


V_3 = ((9* 10^9)(70 * 10^(-9)))/(√(6^2 + 8^2))


V_3 = 63 Volts

Now total potential at this point is given as


V = 45 - 60 + 63 = 48 Volts

User Suresh Chaudhary
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