JEE MAIN - Physics (2022 - 27th July Evening Shift - No. 9)
A charge of $$4 \,\mu \mathrm{C}$$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :
$$1 \,\mu \mathrm{C}$$ and $$3 \,\mu\mathrm{C}$$
$$2 \,\mu \mathrm{C}$$ and $$2\, \mu \mathrm{C}$$
0 and $$4\, \mu\, \mathrm{C}$$
$$1.5 \,\mu \mathrm{C}$$ and $$2.5\, \mu \mathrm{C}$$
Explanation
_27th_July_Evening_Shift_en_9_1.png)
so $$F = {{kq(4 - q) \times {{10}^{ - 12}}} \over {{r^2}}}$$
so Fmax will be at q = 2 $$\mu$$C
Comments (0)
