The spatial distribution of electric field due to charges $(\mathrm{A}, \mathrm{B})$ is shown in figure.…
- $\mathrm{A}$ is $+\mathrm{ve}$ and $\mathrm{B}-\mathrm{ve},|\mathrm{A}|>|\mathrm{B}|$
- $\mathrm{A}$ is $-\mathrm{ve}$ and $\mathrm{B}+\mathrm{ve},|\mathrm{A}|=|\mathrm{B}|$
- Both are +ve but $A>B$
- Both are $-$ ve but $A>B$
Solution
Greater intensity of lines near the charge signifies greater magnitude of charge.
So, $\mathrm{A}$ is positive, $\mathrm{B}$ is negative and $|\mathrm{A}|$ > $|\mathrm{B}|$.
Asked in: JEE Mains - Electrostatics - Test 1