Answer:
a) W = 2.26 × 10⁻²¹ J
b) V = - 0.00706 Volts
Explanation:
Given:
The surface charge density, σ = +5.21 pC/m²
Charge of the particle, q₀ = 3.20 × 10⁻¹⁹
distance, d = 2.44 cm = 0.024 m
Now, the electric field (E) due to the uniformly charged sheet is given as:
[tex]E = \frac{\sigma}{2\epsilon_o}[/tex]
a) The work done is given as:
W = q₀Ed
substituting the values, we get
W = [tex]3.20\times 10^{-19}\frac{5.21\times 10^{-12}}{2\times 8.85\times 10^{-12}}\times 0.024[/tex]
W = 2.26 × 10⁻²¹ J
b) The electric potential (V) at point P is given as:
V = - Ed
substituting the values, we get
V = - ( [tex]\frac{5.21\times 10^{-12}}{2\times 8.85\times 10^{-12}}\times 0.024[/tex])
V = - 0.00706 Volts
