Question

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC moves from the point x= 0.170 m , y= 0 to the point x= 0.250 m , y= 0.250 m .
How much work W is done by the electric force on the moving point charge?
Express your answer in joules. Use k = 8.99×109 N*m^2/ C^2 for Coulomb's constant: k=1/(4*pi*epsilon0)

Answer 1

Answer:

315.7 x 10⁻³ J

Explanation:

Given from the question;

Charge q₁ = 2.30μC, which is stationary

Charge q₂ = -5.0μC, which moves from;

initial point r₁ = [0.170m, 0]  = $\sqrt{0.170^{2} + 0^{2} }$ = $\sqrt{0.170^{2} }$ = 0.170m

final point r₂ = [0.250m, 0.250m] = $\sqrt{0.250^{2} + 0.250^{2} }$ = $\sqrt{0.0625 + 0.0625}$ = $\sqrt{0.125}$ = 0.3536m

Now, remember that the work done (W) on a moving point charge(q) by an electric force is the change in the potential energy (ΔU) of the charge. i.e

W = ΔU

Where;

ΔU = U₁ - U₂

U₁ = potential energy at point r₁

U₂ = potential energy at point r₂

=> W = U₁ - U₂      ----------------------------(i)

Potential energy (U) of between two point charges at a point r is given by;

U = k x q₁ x q₂ / r

Where;

k = 8.99 x 10⁹ Nm²/C²

Therefore, we can write;

U₁ = k x q₁ x q₂ / r₁      and

U₂ = k x q₁ x q₂ / r₂

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

=> W =  (k x q₁ x q₂ / r₁)  -  (k x q₁ x q₂ / r₂)

=> W = k x q₁ x q₂ [$\frac{1}{r_{1} }$ - $\frac{1}{r_{2} }$ ]           --------------------(ii)

Substitute the values of k, q₁, q₂  and r₂ into the equation (ii) as follows;

=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [$\frac{1}{0.170 }$ - $\frac{1}{0.3536 }$]

=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [5.882- 2.828]

=> W = 8.99 x 10⁹ x 2.30 x 10⁻⁶ x 5.00 x 10⁻⁶ x [3.054]

=> W = 315.7 x 10⁻³

=> W = 315.7 mJ

Therefore, the work done on the moving point charge q₂ is 315.7 x 10⁻³ J

Answer 2

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$