# Static electricity – problems and solutions

**Electric force**

1. Point A in an electric field. The magnitude of the electric field at point A = 0.5 NC^{-1}. If a 0.25-C charge placed at point A, then what is the electric force exerted on the charge.

__Known :__

Electric field at point A = 0.5 NC^{-1}

Electric charge at point A = 0.25 C

__Wanted:__ Electric force

__Solution :__

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 × 10^{9} Nm^{2 }C^{–2}, what is the magnitude of electric force experienced by two charges?

__Known :__

Charge 1 (q_{1}) = 5 C

Charge 2 (q_{2}) = 4 C

Distance between charge 1 and charge 2 (r) = 3 meters

Coulomb’s constant (k) = 9 × 10^{9} Nm^{2 }C^{–2}

__Wanted :__ The magnitude of the electric force (F)

__Solution :__

3. Electric charge +q_{1} = 10 μC ; +q_{2} = 20 μC ; and q_{3} are separated, as shown in figure below. The electric force exerted on charge q_{2} = 0, then what is the charge of q_{3}.

__Known :__

Charge 1 (q_{1}) = 10 μC = 10 x 10^{-6 }C

Charge 2 (q_{2}) = 20 μC = 20 x 10^{-6 }C

__Wanted :__ What is the charge of q_{3}

__Solution :__

There are two forces acts on +q_{2}.

The first force is the repulsive force between charge +q_{1 }and charge +q_{2} that is F_{12}, rightward.

The net electric force acts on q_{2} = 0 then q_{3} must negative.

The second force is the attractive force between charge +q_{2} and -q_{3} that is F_{23}, leftward.

Both of these forces act on q_{2}, have the same magnitude but the opposite direction.

Net force acts on +q_{2} = 0.

4. Three charges q_{1}, q, and q_{2} 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.

__Known :__

Charge 1 (q_{1}) = 30 μC = 30 x 10^{-6} C

Charge 2 (q_{2}) = 60 μC = 60 x 10^{-6} C

Charge 3 (q) = 5 μC = 5 x 10^{-6} C

Distance between q_{1} and q = d

Distance between q_{2} and q = 2d

d = 30 cm = 0.3 meters

d^{2 }= (0.3)^{2} = 0.09

Coulomb’s constant (k) = 9 x 10^{9} N m^{2} C^{-2}

__Wanted :__ The magnitude and the direction of the electric force acts on the electric charge

__Solution :__

There are two forces act on q that is F_{1} rightward (q and q_{1} are positive so F_{1} away from q and q_{1}) and F_{2} leftward (q and q_{2} are positive so F_{2} away from q and q_{2}). First calculated F_{1} and F_{2}.

The net force is 7.5 Newton. The direction of the net force = the direction of F_{1}, rightward point to charge q_{2}.

**Electric field**

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

__Known :__

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

__Solution :__

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 Q_{A} = –300 μC, Q_{B }= 600 μC. 1/4 π ε_{0} = 9 × 10^{9 }N m^{2 }C– 2, then what is the magnitude of the electric field at point C produced by both charges.

__Known :__

Distance between charge A and B (r_{AB}) = 4 meters

Distance between point C and charge A (r_{AC}) = 1 meter

Distance between point C and charge B (r_{BC}) = 3 meters

Charge A (q_{A}) = –300 μC = -300 x 10^{-6} C = -3 x 10^{-4} Coulomb

Charge B (q_{B}) = 600 μC = 600 x 10^{-6 }C = 6 x 10^{-4} Coulomb

Constant (k) = 9 × 10^{9 }N m^{2 }C^{–2}

__Wanted :__ Electric field at point C

__Solution :__

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 = E_{A} + E_{B}

E = (27 x 10^{5}) + (6 x 10^{5})

E = 33 x 10^{5 }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.

__Known :__

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/s^{2}

__Wanted :__ Electric field

__Solution :__

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/s^{2}) = 10 x 10^{-6} kg m/s^{2} = 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 q_{1 }= 32 μC and q_{2} = -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 q_{2}. Find x.

__Known :__

Charge 1 (Q_{1}) = 32 μC

Charge 2 (Q_{2}) = -214 μC

Distance of point P from q_{1} = x + 10 cm

The distance of point P from q_{2} = 10 cm

__Wanted:__ x

__Solution :__

E_{1} = electric field produced by charge Q_{1}. The direction of the electric field away from Q_{1} because Q_{1} is positive. E_{2} = electric field produced by Q_{2}. The direction of the electric field point to Q_{2} because Q_{2} is negative.

No net force at point p located at 10 cm from Q_{2 }

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.

__Known :__

Electric force (F) = 0.05 Newton

Electric charge (Q) = +5 × l0^{–6 }Coulomb = 0.000005

__Wanted :__ Electric field at point P

__Solution :__

E = F / Q

E = 0.05 Newton / 0.000005 Coulomb

E = 5 Newton / 0.0005 Coulomb

E = 10,000 Newton/Coulomb

E = 10^{4 }N/C

E = 10^{4 }NC^{-1}