CBSE Class 12-science Answered
A point charge qa = + 100 c is placed at point a (1, 0, 2) m and another point charge qb = +200c isplaced at point b (4, 4, 2) m. Find :(i) magnitude of electrostatic interaction force acting between them(ii) find a f(force on a due to
Asked by raghavsumit9690 | 10 Jul, 2021, 13:30: PM
Electrostatic interaction force F between charges is given as

Where K = 1/(4πεo ) = 9 × 109 N m2 C-2 is Coulomb's constant and εo is permittivity of free space .
Distance between charges is given as d in above expression
d = 

F = ( 9 × 109 × 100 × 200 ) / 25 = 7.2 × 1012 N
----------------------------------
Force on charge qa due to charge qb = 7.2 × 1012 N
Answered by Thiyagarajan K | 10 Jul, 2021, 15:33: PM
Concept Videos
CBSE 12-science - Physics
Asked by gy876934 | 28 Sep, 2024, 07:18: AM
CBSE 12-science - Physics
Asked by ab4300816 | 07 Sep, 2024, 15:06: PM
CBSE 12-science - Physics
Asked by yy8965818 | 07 Jun, 2024, 20:36: PM
CBSE 12-science - Physics
Asked by kanishkg511 | 30 Apr, 2024, 19:25: PM
CBSE 12-science - Physics
Asked by rishabhverma895334 | 01 Mar, 2024, 07:24: AM
CBSE 12-science - Physics
Asked by rameshsanju123 | 08 Feb, 2024, 20:45: PM
CBSE 12-science - Physics
Asked by bhatig9772 | 09 Jan, 2023, 18:48: PM
CBSE 12-science - Physics
Asked by bhagawan5225 | 06 May, 2022, 05:58: AM
CBSE 12-science - Physics
Asked by nonuhasan2 | 28 Nov, 2021, 21:43: PM
CBSE 12-science - Physics
Asked by komalshakatawat18 | 19 Sep, 2021, 17:47: PM