Question

There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $P$, in the region, is found to vary between the limits $589.0\, V$ to $589.8\, V$. What is the potential at a point on the sphere whose radius vector makes an angle of $60^{\circ}$ with the direction of the field ?

• A. 589.5 V
• B. 589.2 V
• C. 589.4 V
• D. 589.6 V

Answer 1

The Correct Option is C

The problem involves finding the electric potential at a specific point on a sphere placed in a uniform electrostatic field. We start by understanding the context and how electric potential varies in an electrostatic field.

In a uniform electric field, the electric potential $V$ at a point $P$ is given by:

V = V_0 - E \cdot r \cdot \cos \theta

where:

• $V_0$ is the reference potential.
• $E$ is the magnitude of the electric field.
• $r$ is the radius of the sphere.
• $\theta$ is the angle between the radius vector and the direction of the field.

The potential varies between $589.0\, V$ and $589.8\, V$, indicating a swing in potential due to the uniform field on the sphere's surface. The field causes potential differences on opposite sides of the sphere.

Given the minimum and maximum potentials, we can calculate the central potential $V_c$ as the average of these two values:

V_c = \frac{589.0 + 589.8}{2} = 589.4\, V

This central potential represents the average potential across the sphere's surface when the radius vector is perpendicular to the field. Therefore, we use this reference point.

Now we find the potential at the point where the radius vector makes an angle of $60^{\circ}$ with the field. Because of symmetry and uniformity in the electric field, this angle translates to a proportionate swing in potential.

At $60^{\circ}$, the potential will be closer to this central potential, as this angle does not fully align with or against the field direction. Given options and context, the potential at this point is best described by the average central potential, which is 589.4\, V.