81 Views

two charges 3.8 into 10 to the power 5 cm and 5.5 into 10 power 4 Siya place to get the distance and 10 cm each and other find the the value of electricity force acting between them


Navnath Maynale 24th Aug, 2020
Answer (1)
Prabhav Sharma 24th Aug, 2020

Hello,

Here, Q1 = 3.8 * 10^5 and Q2 = 5.5 * 10^4

Distance between them, r = 10*10^-2

According to Coloumb's law, force acting between two bodies having charged Q1 and Q2 placed r distance apart is given by:

F = 1/4(pie)(Epsilon) * (Q1*Q2)/r^2

Substituting the values in the above equation, we get:

F = (1/4 * 3.14 * 8.85 * 10^-12) * (3.8 * 10^5 * 5.5 * 10^4) / (10*10^-2)^2

F = 1.88 * 10^22 N (Answer)

Related Questions

Amity University | Journalism...
Apply
Ranked amongst top 3% universities globally (QS Rankings)
Sanskaram University MJMC Adm...
Apply
100+ Industry collaborations | 10+ Years of legacy
Graphic Era (Deemed to be Uni...
Apply
NAAC A+ Grade | Among top 100 universities of India (NIRF 2024) | 40 crore+ scholarships distributed
MAHE Manipal Liberal Arts Adm...
Apply
Accorded Institution of Eminence by MoE, Govt. of India | NAAC A++ Grade | Ranked #4 India by NIRF 2024
UPES Dehradun -Media and Mass...
Apply
Ranked #46 Among Universities in India by NIRF | 1950+ Students Placed | 91% Placement | Last Date to Apply: 29th May
Sanskaram University BJMC Adm...
Apply
100+ Industry collaborations | 10+ Years of legacy
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books