Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me understand! Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes

Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between the charges becomes

  • Option 1)

    \frac{F}{16}\;

  • Option 2)

    \; \frac{9F}{16}\; \;

  • Option 3)

    \; F\; \;

  • Option 4)

    \; \frac{15}{16}F

 

Answers (1)

As we learned

Coulomb's Law -

The force of attraction or repulsion is directly proportional to product of their two charges and inversly proportional to square distance between them.

-

 

 

Initially  F=k\frac{Q^{2}}{r^{2}}\; \; \; \; \; \; \; Finally\; F'=\frac{k.(\frac{Q}{4})^{2}}{r^{2}}=\frac{F}{16}

 


Option 1)

\frac{F}{16}\;

Option 2)

\; \frac{9F}{16}\; \;

Option 3)

\; F\; \;

Option 4)

\; \frac{15}{16}F

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA Exam Changes Live: Reforms in NEET UG, JEE Main, CUET; test centres, question papers, biometric data
anam-khan

NTA to not hold recruitment exams; will be restructured in 2025: Dharmendra Pradhan
anam-khan

JEE Main 2025: BTech ECE remains among top engineering choices; NIT cut-offs

Latest Question