When two identical sphere come in contact with each other, the total charge on them is equally distribute.
\(\frac{kQ^2}{d^2}\)=F
F=\(\frac{k9Q^2}{16×\frac{d^2}{4}}−\frac{k3Q^2}{8×\frac{d^2}{4}}\)
=\(\frac{9kQ^2}{4d^2}−\frac{3kQ^2}{2d^2}\)
=\(\frac{kQ^2}{d^2}[\frac{9}{4}−\frac{3}{2}]\)
=\(\frac{6}{8}F\)
=\(\frac{3}{4}F\)
So, the correct option is (B): \(\frac{3}{4}F\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
A spherical mirror is a mirror which has been cut out of a spherical surface.
There are two kinds of spherical mirrors:
Concave mirrors are also called converging mirrors, because in these types of mirrors, light rays converge at a point after impact and reflect back from the reflective surface of the mirror.
The convex mirror has a reflective surface that is curved outward. Regardless of the distance between the subject and the mirrors, these mirrors are "always" virtual, upright and reduced.