Logo
Jeeify
MCQEasyJEE 2026Electric Field & Field Lines

JEE Physics 2026 Question with Solution

A simple pendulum has a bob with mass mm and charge qq. The pendulum string has negligible mass. When a uniform and horizontal electric field E⃗\vec{E} is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is _____.

(gg: acceleration due to gravity)

  • A

    m2g2−q2E2\sqrt{m^2 g^2 - q^2 E^2}

  • B

    m2g2+q2E2\sqrt{m^2 g^2 + q^2 E^2}

  • C

    mg+qEmg + qE

  • D

    mg−qEmg - qE

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A simple pendulum bob has mass mm and charge qq. A uniform horizontal electric field E⃗\vec{E} is applied, and gravitational acceleration is gg.

Find: The final tension TT in the string when the pendulum is in equilibrium.

At equilibrium, the bob experiences two perpendicular forces:

  • Gravitational force mgmg vertically downward
  • Electric force qEqE horizontally

The tension balances the resultant of these two forces.

T=(mg)2+(qE)2T = \sqrt{(mg)^2 + (qE)^2}

Therefore,

T=m2g2+q2E2T = \sqrt{m^2 g^2 + q^2 E^2}

Therefore, the correct option is B.

Force Balance Explanation

Given: The bob is under the action of gravity and a horizontal electric force.

Find: The magnitude of the string tension at equilibrium.

Step 1: Identify forces acting on the bob. The bob experiences:

  • Gravitational force mgmg vertically downward
  • Electric force qEqE horizontally

Step 2: Condition of equilibrium. At equilibrium, the string tension must balance the vector sum of these two mutually perpendicular forces.

Using Pythagoras theorem,

T=(mg)2+(qE)2T = \sqrt{(mg)^2 + (qE)^2}

So,

T=m2g2+q2E2T = \sqrt{m^2 g^2 + q^2 E^2}

Hence, the final tension is m2g2+q2E2\sqrt{m^2 g^2 + q^2 E^2}.

Common mistakes

  • Treating mgmg and qEqE as forces in the same direction and writing T=mg+qET = mg + qE. This is wrong because gravity acts vertically while electric force acts horizontally. Instead, take the resultant using perpendicular vector addition.

  • Using T=mg−qET = mg - qE or m2g2−q2E2\sqrt{m^2 g^2 - q^2 E^2}. This is incorrect because the forces do not oppose each other along one line. Instead, use Pythagoras theorem for the two perpendicular forces.

Practice more Electric Field & Field Lines questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Electric Field & Field Lines

Related questions