Related Problems with Solution
Problem 1 : A uniformly charged infinite plane has a charge density of $$(2 , \mu\text{C/m}^2).$$ Calculate the electric field at a point 3 meters above the plane.
Solution :
For an infinite uniformly charged plane, the electric field above the plane is constant and perpendicular to the plane.
Electric field (E) due to a uniformly charged infinite plane: $$[E = \frac{\sigma}{2\epsilon_0}]$$
Where:
- σ is the surface charge density $$((2 \times 10^{-6} , \text{C/m}^2))$$
- ε0 is the permittivity of free space $((8.85 \times 10^{-12} , \text{C}^2/\text{N}\cdot\text{m}^2))$$
Substitute the values to calculate (E): $$[E = \frac{2 \times 10^{-6} , \text{C/m}^2}{2 \cdot 8.85 \times 10^{-12} , \text{C}^2/\text{N}\cdot\text{m}^2} = 112.99 , \text{N/C}]$$
So, the electric field 3 meters above the plane is 112.99 N/C directed upward.
