Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:
- A$\frac{3}{4} F$
- B$\frac{4}{3} F$
- C$\frac{9}{16} F$
- D$\frac{16}{9} F$
Explore More
Similar Questions
Electric charges of $1 \mu C, -1 \mu C$ and $2 \mu C$ are placed in air at the corners $A, B$ and $C$ respectively of an equilateral triangle $ABC$ having a side length of $10 \ cm$. The resultant force on the charge at $C$ is: (in $N$)
Difficult
View SolutionThree equal charges $q_1, q_2$ and $q_3$ are placed on the three corners of a square of side $a$. If the force between $q_1$ and $q_2$ is $F_{12}$ and that between $q_1$ and $q_3$ is $F_{13}$,then the ratio of magnitudes $\left(\frac{F_{12}}{F_{13}}\right)$ is
Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:
Two identical conducting spheres $A$ and $B$ having charges $+q$ and $-q$ respectively are kept at a distance $d$ apart and experience a Coulombian force $F$ between them. If $50\%$ of the charge is transferred from sphere $B$ to $A$,then the new Coulombian force between them is . . . . . . .
Write some important points for the vector form of Coulomb's law.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo