A small metallic sphere is held in a non-uniform electrostatic field a

A small metallic sphere is held in a non-uniform electrostatic field as shown in the figure. Also shown are two points P and Q on the surface of the sphere. Which one of the following statements is true for the sphere?

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Sphere does not experience a force in the field.

Surface charge density at point P is positive while at point Q is negative.

Sphere experiences a force towards the left.

Sphere experiences a force towards the right.

Answer is Right!

Answer is Wrong!

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UPSC Geoscientist – 2024

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When a conductor is placed in a non-uniform electrostatic field, charges are induced on its surface. Due to the non-uniformity of the field, the forces on the induced positive and negative charges do not cancel out, resulting in a net force on the conductor.

A conductor in an electrostatic field is an equipotential surface, and the electric field lines are perpendicular to its surface. In a non-uniform field (indicated by varying density of field lines), the side facing the stronger field region (denser lines, like point P) will accumulate induced charge opposite to the field source, and the opposite side (sparser lines, like point Q) will accumulate the same type of charge as the source. The forces on these induced charges are F = qE. Since the field strength E is non-uniform, the force on the induced charge in the stronger field region will be greater.

The diagram shows field lines entering the sphere predominantly from the left, indicating a stronger field on the left (around P) and a weaker field on the right (around Q). Induced negative charge will accumulate at P, and induced positive charge at Q (assuming the field source is effectively positive and to the left, causing field lines to point rightwards). The force on the negative charge at P (F_P) will be towards the left (opposite to the field direction). The force on the positive charge at Q (F_Q) will be towards the right (in the field direction). Since the field is stronger at P, |F_P| > |F_Q|. The net force F_net = F_P + F_Q is thus directed towards the left (towards the region of stronger field). Statement B is incorrect regarding charge signs.

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