Electric field at point A = 0.5 NC-1
Electric charge at point A = 0.25 C
Wanted: Electric force
F = q E
F = (0.25 C)(0.5 NC-1)
F = 0.125 N
2. Two charges 5 C and 4 C are separated by a distance of 3 meters. If Coulomb constant is 9 × 109 Nm2 C–2, what is the magnitude of electric force experienced by two charges?
Charge 1 (q1) = 5 C
Charge 2 (q2) = 4 C
Distance between charge 1 and charge 2 (r) = 3 meters
Coulomb’s constant (k) = 9 × 109 Nm2 C–2
Wanted : The magnitude of the electric force (F)
3. Electric charge +q1 = 10 μC ; +q2 = 20 μC ; and q3 are separated, as shown in figure below. The electric force exerted on charge q2 = 0, then what is the charge of q3.
Charge 1 (q1) = 10 μC = 10 x 10-6 C
Charge 2 (q2) = 20 μC = 20 x 10-6 C
Wanted : What is the charge of q3
There are two forces acts on +q2.
The first force is the repulsive force between charge +q1 and charge +q2 that is F12, rightward.
The net electric force acts on q2 = 0 then q3 must negative.
The second force is the attractive force between charge +q2 and -q3 that is F23, leftward.
Both of these forces act on q2, have the same magnitude but the opposite direction.
Net force acts on +q2 = 0.
4. Three charges q1, q, and q2 are in a line. If q = 5.0 μC and d = 30 cm, then what is the magnitude and direction of the electric force acts on the charge q.
Charge 1 (q1) = 30 μC = 30 x 10-6 C
Charge 2 (q2) = 60 μC = 60 x 10-6 C
Charge 3 (q) = 5 μC = 5 x 10-6 C
Distance between q1 and q = d
Distance between q2 and q = 2d
d = 30 cm = 0.3 meters
d2 = (0.3)2 = 0.09
Coulomb’s constant (k) = 9 x 109 N m2 C-2
Wanted : The magnitude and the direction of the electric force acts on the electric charge
There are two forces act on q that is F1 rightward (q and q1 are positive so F1 away from q and q1) and F2 leftward (q and q2 are positive so F2 away from q and q2). First calculated F1 and F2.
The net force is 7.5 Newton. The direction of the net force = the direction of F1, rightward point to charge q2.
5. A point charge q is at the point P in the electric field produced by the charge (+) so that it experiences a force of 0.05 N in the direction towards the charge. If the magnitude of the electric field at point P is 2 x 10-2 NC-1, then what is the magnitude and type of charge that causes the electric field.
Electric force (F) = 0.05 N
Electric field (E) = 2 x 10 –2 NC –1 = 0.02 NC –1
Wanted: The magnitude and type of electric charge
The electric charge is calculated using a formula that expresses the relationship between electric force (F), the electric field (E) and electric charge (q):
F = q E
q = F / E = 0.05 N / 0.02 NC –1 = 2.5 Coulomb
The charge q experiences an electric force in the direction to charge (+) which produces an electric field, so the charge q is negative.
6. Charge A and B are separated by a distance of 4 meters. Point C is between both charges, 1 meter from point A. If QA = –300 μC, QB = 600 μC. 1/4 π ε0 = 9 × 109 N m2 C– 2, then what is the magnitude of the electric field at point C produced by both charges.
Distance between charge A and B (rAB) = 4 meters
Distance between point C and charge A (rAC) = 1 meter
Distance between point C and charge B (rBC) = 3 meters
Charge A (qA) = –300 μC = -300 x 10-6 C = -3 x 10-4 Coulomb
Charge B (qB) = 600 μC = 600 x 10-6 C = 6 x 10-4 Coulomb
Constant (k) = 9 × 109 N m2 C–2
Wanted : Electric field at point C
Electric field produced by charge A at point C :
Charge A is negative so that the direction of the electric field point to charge A and away from charge B (to left).
The electric field produced by charge B on point C :
Charge B is positive so that the direction of electric field away from charge B and point to charge A (to left).
The resultant of the electric field at point A :
E = EA + EB
E = (27 x 105) + (6 x 105)
E = 33 x 105 N/C
The direction of the electric field point to charge A and away from charge B (to left).
7. A 1-mg dust float in the air. If the charge of the dust is 0.5 μC and acceleration due to gravity is 10 m/s2, determine the magnitude of the electric field that supports dust.
Mass of dust (m) = 1 mg = 1 x 10-6 kg
Charge of dust (q) = 0.5 μC = 0.5 x 10-6 C
Acceleration due to gravity (g) = 10 m/s2
Wanted : Electric field
w = m g
w = weight of dust, m = mass of dust, g = acceleration due to gravity
The force of gravity acts on dust :
w = m g = (1 x 10-6 kg)(10 m/s2) = 10 x 10-6 kg m/s2 = 10 x 10-6 Newton
The equation of the electric field :
E = F/q
E = electric field, F = electric force, q = electric charge
Substitute F in the equation of electric field with w in the equation of weight :
E = F/q = w/q
E = (10 x 10-6 N) / (0.5 x 10-6 C)
E = 10 N / 0,5 C
E = 20 N/C
8. Two charges q1 = 32 μC and q2 = -214 μC are separated by a distance of x, as shown in figure below. No electric field at point P located at 10 cm from q2. Find x.
Charge 1 (Q1) = 32 μC
Charge 2 (Q2) = -214 μC
Distance of point P from q1 = x + 10 cm
The distance of point P from q2 = 10 cm
E1 = electric field produced by charge Q1. The direction of the electric field away from Q1 because Q1 is positive. E2 = electric field produced by Q2. The direction of the electric field point to Q2 because Q2 is negative.
No net force at point p located at 10 cm from Q2
Use quadratic formula :
9. A charged point q is at the point P in the electric field produced by the charge (+), thus having a force of 0.05 N. If the charge is +5 × l0–6 Coulomb, then what is the magnitude of the electric field at point C.
Electric force (F) = 0.05 Newton
Electric charge (Q) = +5 × l0–6 Coulomb = 0.000005
Wanted : Electric field at point P
E = F / Q
E = 0.05 Newton / 0.000005 Coulomb
E = 5 Newton / 0.0005 Coulomb
E = 10,000 Newton/Coulomb
E = 104 N/C
E = 104 NC-1