A particle of charge +8.2 μC is released from rest at the point x = 70 cm on an x axis. The particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 34 cm if (a)Q = +31 μC and (b)Q = –31 μC?

(a) 1.07J

(b) 3.09J

In an electric field, the change in potential energy (ΔU) of a particle results in a corresponding reversed change in kinetic energy (ΔK). i.e

ΔU = -ΔK

But;

ΔU = U₂ - U₁

ΔK = K₂ - K₁

Equation (i) can be re-written as;

U₂ - U₁ = - ( K₂ - K₁)

U₂ - U₁ =  K₁ - K₂       --------------------(ii)

Where;

U₁ = potential energy between a particle of charge q and a point charge Q, at some initial position r₁

U₂ = potential energy  between a particle of charge q and a point charge Q, at some later position r₂

K₁ = kinetic energy  between a particle of charge q and a point charge Q, at some initial position r₁

K₂ = kinetic energy  between a particle of charge q and a point charge Q, at some later position r₂

The potential energy U  between a particle of charge q and a point charge Q, at some point, r, is given as;

U = k x q x Q / r      ---------------(iii)

Where;

k = constant = 8.99 x 10⁹ Nm²/C²

Using equation (iii) we can write U₁ and U₂ as;

U₁ = k x q x Q / r₁  

U₂ = k x q x Q / r₂

Also,

K₁ = 0      [since the particle starts from rest]

K₂ = unknown = kinetic energy at the instant the particle has moved 34cm

Substitute these values of U₁, U₂, K₁ and K₂ into equation (i) as follows;

( k x q x Q / r₂ ) - ( k x q x Q / r₁ ) = 0 - K₂

( k x q x Q / r₂ ) - ( k x q x Q / r₁ ) =  - K₂  

k x q x Q [ - ]  =  - K₂           -------------------(iv)

From the question;

r₁ = 70cm = 0.7m

q = +8.2μC = +8.2 x 10⁻⁶ C

(a) When Q is +31μC = +31 x 10⁻⁶ C

Since the two charges are positive, the particle will move 34cm further away (repulsion) i.e from 70cm to 104cm. Therefore;

r₂ = 70 + 34 = 104cm = 1.04m

Substitute the value of Q, q, r₁, r₂ and k into equation (iv) as follows;

8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [ - ] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [0.96 - 1.43] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ [-0.47] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x 31 x 10⁻⁶ x  [-0.47] = -K₂

-1.07 = -K₂

K₂ = 1.07J

Therefore, the kinetic energy at that instant is 1.07 J

(b) When Q is -31μC = -31 x 10⁻⁶ C

Since the two charges have different polarities (one is positive and the other is negative), the particle will move 34cm closer (attraction) i.e from 70cm to 36cm. Therefore;

r₂ = 70 - 34 = 36cm = 0.36m

Substitute the value of Q, q, r₁, r₂ and k into equation (iv) as follows;

8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [ - ] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [2.78 - 1.43] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ [1.35] = -K₂

8.99 x 10⁹ x 8.2 x 10⁻⁶ x -31 x 10⁻⁶ x  1.35 = -K₂

-3.09 = -K₂

K₂ = 3.09 J

Therefore, the kinetic energy at that instant is 3.09 J

the answer would be 32*f