Q&A - Ask Doubts and Get Answers
Q

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)
75 Views
V Vakul

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

Exams
Articles
Questions