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

Explanation:

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 objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.

The answer is 12, for the points