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The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.

Sagot :

Numenius

Answer:

The potential at a distance of 0.9 m is 266.67 V.

Explanation:

Charge = Q

Potential is 400 V at a distance 0.6 m .

Let the potential is V at a distance 0.9 m.

Use the formula of potential.

[tex]V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V[/tex]

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