A positive charge of 3 × 10–8C is in a uniform electric field of field strength 300 000V m–1. How much work must be done on the charge in order to move it a distance of 4.0 mm in the opposite direction to the direction of the field?​

Answers (1)
S safeer

Force F=qE

F=2*10-8*300000=6*10-3N

W=force*displacement=-6*10-3*4*10-3=24*10-6J

Negative sign indicates that work done is against the force

Preparation Products

Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Test Series NEET May 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions