JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
$$C_1$$ and $$C_2$$ are two hollow concentric cubes enclosing charges $$2Q$$ and $$3Q$$ respectively as shown in figure. The ratio of electric flux passing through $$C_1$$ and $$C_2$$ is
By Gauss law,
$$\Phi=\frac{Q_{enclosed}}{\epsilon_0}$$For inner cube $$C_1$$, only charge 2Q is enclosed.
$$\Phi_1=\frac{2Q}{\epsilon_0}$$
For outer cube $$C_2$$, both charges 2Q and 3Q are enclosed.
$$\Phi_2=\frac{2Q+3Q}{\epsilon_0}=\frac{5Q}{\epsilon_0}$$Therefore,
$$\frac{\Phi_1}{\Phi_2}=\frac{2}{5}$$
So ratio is
$$2:5$$
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
