Question

2.) A +12 uC charge is located at the origin (x-o cm) and a-4 uC charge is located at x-S cm the voltage at the point A 10 cm) and the yoltage at the point B (x 3 cm), and then determine the work done BY the electric field in moving a -6 uC charge from A to B. Will you have to force this to happen or will

Answer 1

4 uC +12 uC x= 10 cm x=3cm x=8cm x-0

The formula for calculating electric potential due to point charges is

2 e互 1グ

a)

Therefore the voltage at point A(x=10cm) -

V_tA = K Q₁ / R₁ + K Q₂ / R₂

where -

K = (nearby) 9 x 10⁹ NC/m²

Q₁ = +12 uC = 12 x 10^-6 C

Q₂ = -4 uc = -4 x 10^-6 C

R₁ = (10 - 0) cm = 10 x 10^-2 m

R₂ = (10 - 8) cm = 2 x 10^-2 m

By putting these values in the equation we get -

V_tA = - 719,004 V

b)

Therefore the voltage at point B(x=3cm) -

V_tA = K Q₁ / R₁ + K Q₂ / R₂

where -

K = (nearby) 9 x 10⁹ NC/m²

Q₁ = +12 uC = 12 x 10^-6 C

Q₂ = -4 uc = -4 x 10^-6 C

R₁ = (10 - 3) cm = 3 x 10^-2 m

R₂ = (8 - 3) cm = 5 x 10^-2 m

By putting these values in the equation we get -

V_tA = 2,876,016 V

c)

Formula for calculating Work Done from Potential Difference is -

Fd W Ed=-=-=AV For constant electric field. Nm Joule Units: m- = Volts

Since,

V = W / Q

=> W = V x Q

where -

V = Potential difference

W = Work Done

Q = Charge of the particle

Since charge is being moved from A to B, therefore

Potential Difference =V = V_tB - V_tA = 2,876,016 - (-719,004) = 2,157,012 V

Q = - 6 uC = -6 x 10^-6 C

=> V = 2,157,012 x (-6 x 10^-6 ) = - 12.942 J

d)

It will happen spontaneously.

The direction of field is from B to A(Decreasing potential of field is the direction of electric field). It means point B behaves like a + charge and point A behaves like a - charge.

Therefore a negative charge will travel spontaneously in opposite direction of the electric field ie A to B.