Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
The work of electric field done during the displacement of a negatively charged particle towards a fixed positively charged particle is 9J. As a result the distance between the charges has been decreased by half. What work is done by the electric field over the first half of this distance?
Ans: 3J
How we are getting this answer plz explain
given:
kQq/(x/2) - kQq/x =9 where x is the intial distance. . . .
or kQq/x = 9J
to find is
KQq/(3/4*x) - kQq/x .... as the rdistance b/w the particles becomes 3/4*x as the particle approches by a quater*x
therefore it is kQq/3x =9/3 =3
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !