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Four identical point charges are arranged at the corners of a square [Hint: Draw a figure]. The electric field E and potential V at the centre of the square are

(a) \(E = 0\), \(V = 0\).

(b) \(E = 0\), \(V \ne 0\).

(c) \(E \ne 0\), \(V \ne 0\).

(d) \(E \ne 0\), \(V = 0\).

(e) \(E = V\) regardless of the value.

Short Answer

Expert verified

The correct answer is option (b) \(E = 0\), \(V \ne 0\).

Step by step solution

01

Understanding the electric field

The presence of an electric field is only due to the difference in electric potential between two points in space. The electric field is the gradient of electric potential. It inversely depends on the distance of a point from a charge.

02

Evaluation of the values of electric field and electric potential

The electric field at any point due to a point charge is given by,

\(\overrightarrow E = \frac{{kQ}}{{{r^2}}}\hat r\)

Here, k is the Coulomb’s constant, Q is the charge and r is the distance.

The electric field is a vector quantity. Since all the charges placed at the corners of a square are identical, the magnitude of electric field at the centre of the square due to all four charges is the same but in different directions. At the centre of the square, they add up to give a net electric field zero.

The electric potential is given as:

\(V = \frac{{kQ}}{r}\)

Electric potential is a scalar quantity. Therefore, the electric potential due to four identical charges has some value at the centre of the square.

Therefore, at the centre,\(E = 0;\;V \ne 0\)

Thus, the correct option is (b).

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Most popular questions from this chapter

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