Question

1. An electric charge inside a sphere (r=0.81m) has a magnitude of +4.8x10−18𝐶. Find the electric flux through this surface.

2. An electric field (E=4.5N/C) passes through a flat surface area (0.30 m) with an angle of 45˚. What will be the magnitude of the electric flux?

3. A closed surface which encloses a charge has an electric flux of 15.8𝑁.𝑚 2 . 𝐶 What is the magnitude of the charge?

Answer 1

1. The electric flux through the Gaussian surface

$$\Phi=\oint EdA=\frac{q}{\epsilon_0}$$ $$\Phi=\frac{4.8*10^{−18}}{8.85*10^{-12}}=5.4*10^{-7}\:\rm N\cdot m^2/C$$

2.

$$\Phi=EA\cos\theta\\=4.5*0.30*\cos45^\circ=0.95\:\rm N\cdot m^2/C$$

3.

$$q=\Phi*\epsilon_0=15.8*8.85*10^{-12}=1.40*10^{-10}\:\rm C$$