A positive charge of +2 μC is placed at point A in an electric field with a magnitude of 5 N/C, directed towards the positive x-axis. The electric field extends along the x-axis from -2 m to +5 m.
(a) Calculate the electric force exerted on the positive charge at point A.
(b) Calculate the work done by the electric field on the positive charge as it moves from point A to point B, located at +4 m on the x-axis.
(c) Determine the potential difference between points A and B.
(d) If the positive charge is released from point A, calculate its speed when it reaches point B, assuming no other external forces act on it.
Provide your answers to each part with step-by-step explanations.
(a) To calculate the electric force exerted on the positive charge at point A, we use the equation:
F = q * E
where F is the force, q is the charge, and E is the electric field strength.
Given: q = +2 μC = 2 * 10^(-6) C E = 5 N/C
Substituting the values:
F = (2 * 10^(-6) C) * (5 N/C)
F = 10 * 10^(-6) N
F = 10 μN
Therefore, the electric force exerted on the positive charge at point A is 10 μN.
(b) To calculate the work done by the electric field on the positive charge as it moves from point A to point B, we use the equation:
W = q * ΔV
where W is the work done, q is the charge, and ΔV is the potential difference.
Given: q = +2 μC = 2 * 10^(-6) C ΔV = V_B - V_A = (E * d) - (E * 0) = E * d d = (4 - 0) m = 4 m E = 5 N/C
Substituting the values:
W = (2 * 10^(-6) C) * (5 N/C) * 4 m
W = (10 * 10^(-6) C) * (5 N/C) * 4 m
W = 200 * 10^(-6) N * m
W = 200 * 10^(-6) J
W = 0.2 mJ
Therefore, the work done by the electric field on the positive charge as it moves from point A to point B is 0.2 mJ.
(c) To determine the potential difference between points A and B, we use the equation:
ΔV = V_B - V_A = E * d
Given: ΔV = V_B - V_A ΔV = E * d E = 5 N/C d = (4 - 0) m = 4 m
Substituting the values:
ΔV = (5 N/C) * 4 m
ΔV = 20 N * m / C
Therefore, the potential difference between points A and B is 20 V.
(d) If the positive charge is released from point A, its initial potential energy will be converted into kinetic energy as it moves towards point B. Assuming no other external forces act on it, we can use the conservation of energy to calculate its speed when it reaches point B.
E_initial = E_final
PE_initial + KE_initial = PE_final + KE_final
0 + 0 = 0 + KE_final
KE_final = 0.2 mJ
Since KE = (1/2)mv^2, we can rearrange the equation to solve for v:
KE_final = (1/2)mv^2
0.2 mJ = (1/2)m(v^2)
v^2 = (2 * 0.2 mJ) / m
v^2 = (0.4 mJ) / m
To find m, we can use the equation q = m * v:
2 * 10^(-6) C = m * v
Rearranging, we get:
m = (2 * 10^(-6) C) / v
Substituting this into the previous equation:
v^2 = (0.4 mJ) / [(2 * 10^(-6) C) / v]
v^2 = (0.4 mJ * v) / (2 * 10^(-6) C)
v^2 = (0.4 * 10^(-3) J * s) / (2 * 10^(-6) C)
v^2 = 0.2 J * s / C
Since 1 J * s / C = 1 V, we can simplify:
v^2 = 0.2 V
v = sqrt(0.2 V)
v ≈ 0.447 V
Therefore, the speed of the positive charge when it reaches point B is approximately 0.447 m/s.