Electric field equation

3 questions about Electric field equation

1. A conducting ball with a radius of 10 cm has an electric charge of 500 μC. Points A, B, and C lie in line with the center of the ball at a distance of 12 cm, 10 cm and 8 cm respectively from the center of the ball. Calculate the electric field strength at points A, B, and C!

Known:

The radius of the conducting ball (R) = 10 cm = 0.1 m

Electric charge (q) = 500 μC = 500 x 10^-6 C

r_A = 12 cm = 0,12 m

r_B = 10 cm = 0,1 m

r_C = 8 cm = 0,08 m

Coulomb constant (k) = 9 x 10⁹

Wanted: The electric field strength at point A (E_A), at point B (E_B) and at point C (E_C)

Solution:

a) The electric field strength at point A

E_A = k q / r_A² = (9 x 10⁹)(500 x 10^-6) / (0,12)² = (4500 x 10³) / 0,0144 = 312500 x 10³ = 3,125 x 10⁸ N/C

b) The electric field strength at point B

E_B = k q / r_B² = (9 x 10⁹)(500 x 10^-6) / (0,1)² = (4500 x 10³) / 0,01 = 450.000 x 10³ = 4,5 x 10⁸ N/C

c) The electric field strength at point C

E_C = 0 for being in the ball.

2. If a test charge of 4 nC is placed at a point, the charge experiences a force of 5 × 10 – 4 N. What is the magnitude of the electric field E at that point?

Known:

Test electric charge (q) = 4 nC = 4 x 10^-9 Coulomb

Electric force (F) = 5 × 10^-4 N

Wanted: The magnitude of the electric field (E)

Solution:

E = F / q = (5 × 10^-4) / (4 x 10^-9) = 1,25 x 10⁵ N/C

3. Two charges q_B = 12 μC and q_C = 9 μC are placed at the vertices of a right triangle as in Fig. Determine the electric field strength felt at point A!

Known:

Charge at point B (qB) = 12 μC = 12 x 10^-6 C

Charge at point C (qC) = 9 μC = 9 x 10^-6 C

Coulomb constant (k) = 9 x 10⁹

r_AC = 4 cm = 0,04 m

r_AB = 3 cm = 0,03 m

Wanted: electric field strength at point A

Solution:

E_AC = k q / r² = (9 x 10⁹)(9 x 10^-6) / (0,04)² = 81 x 10³ / 0,0016 = 5,0 x 10⁷ N/C

E_AB = k q / r² = (9 x 10⁹)(9 x 10^-6) / (0,03)² = 81 x 10³ / 0,0009 = 9,0 x 10⁷ N/C