Two point charges, q1 = 10 x10^-8 C and q2 = -2 x 10^-8 C are separated by distance of 60 cm in air. (a) Find at what distance from the 1st charge, q1, would the electric potential be zero Also calculate the electrostatic potential energy of the system (b)also calculaye the electrostatic potential energy of the system
Answers
Given:
Two point charges, q1 = 10 x10^-8 C and q2 = -2 x 10^-8 C are separated by distance of 60 cm in air.
To find:
(a) Find at what distance from the 1st charge, q1, would the electric potential be zero
(b)also calculaye the electrostatic potential energy of the system
Solution:
From given, we have,
Two point charges, q1 = 10 x10^-8 C and q2 = -2 x 10^-8 C are separated by distance of 60 cm in air.
⇒ q₁ = 10 × 10^{-8} C
⇒ q₂ = -2 × 10^{-8} C
⇒ d = 60 cm = 0.6 m
Let AB be the distance between the charges.
Let "O" be a point at which the electric potential is zero.
Let AO = x and OB = 0.6 - x
So, we have,
Potential O due to charge q₁ = Kq₁/AO
Potential O due to charge q₂ = Kq₂/BO
As potential at O is 0, so we get,
Kq₁/AO + Kq₂/BO = 0
Kq₁/AO = - Kq₂/BO
q₁/AO = - q₂/BO
(10 × 10^{-8})/x = - (-2 × 10^{-8})/ (0.6 - x)
10/x = - 2/(0.6 - x)
2x = 6 - 10x
x = 6/12 = 0.5
Therefore, AO = 0.5 m
(a) The distance from the 1st charge, q1, would the electric potential be zero is 0.5 m
In order to find the potential energy of the system, we use the formula,
U = K q₁q₂/d
U = 9 × 10^9 × ((10 × 10^{-8}) (-2 × 10^{-8}) / 0.6
U = -18 × 10^{-6}/0.6
U = -3 × 10^{-5} J
(b) The electrostatic potential energy of the system is -3 × 10^{-5} J