What is the net force and its direction that the charges at the vertices A and C of the right triangle ABC exert on the charge in vertex B?

__Solution to Problem 1:__

Let F_{AB} be the force of repulsion exerted by the charge at A on the charge at B and F_{CB} be the force exerted by the charge at point C on the charge at point B. The diagram below shows the direction of these two forces. We first use Coulomb’s law (F = k q1 q2 / r^{2}) to find the magnitude of these two forces

|F_{AB}| = k (7 × 10^{-6})((2× 10^{-6}) / (4 × 10^{-2})^{2} = 14 × 10^{-12} k / (16 × 10^{-4}) = 0.875 k × 10^{-8}

|F_{CB}| = k (2× 10^{-6})(2× 10^{-6}) / (2 × 10^{-2})^{2} = k × 10^{-8}

We now use Pythagora’s theorem to find the magnitude of the resultant force F = F_{AB} + F_{CB} (a vector sum)

|F| = √(|F_{AB}|^{2} + |F_{CB}|^{2}) = k × 10^{-8} √( 0.875 ^{2} + 1^{2} ) = 9.00 × 10^{9} × 10^{-8} √( 0.875 ^{2} + 1^{2} ) = 1.20 × 10^{2} N

θ = arctan(|F_{CB}|/ |F_{AB}| ) = arctan( k × 10^{-8} / 0.875 k × 10^{-8}) = 48.8°

Problem 2:

A positive charge q exerts a force of magnitude – 0.20 N on another charge – 2q. Find the magnitude of each charge if the distance separating them is equal to 50 cm.

__Solution to Problem 2:__

The force that q exert on 2q is given by Coulomb’s law:

F = k (q) ( – 2q) / r^{2} , r = 0.5 m , F = – 0.20 N ,

– 0.2 = – 2 q^{2} k / 0.5^{2}

q^{2} = 0.2 × 0.5^{2} / (2 k)

q = √ [ (0.2 × 0.5^{2} / (2 × 9 × 10^{9}) ] = 1.66 × 10^{-6} C

q = 1.66 × 10^{-6} C , -2 q = -3.23 × 10^{-6} C

Problem 3:

Two identical objects, separated by a distance d, with charges equal in magnitude but of opposite signs exert a force of attraction of – 2.5 N on each other. What force do these objects exert on each other if the distance between them becomes 2d?

__Solution to Problem 3:__

Let the two charges be q and -q. The magnitude of the force that q and -q, separated by a distance d, exert on each other is given by Coulomb’s law:

F = k (q) (- q) / d^{2} = – k q^{2} / d^{2} = – 2.5 N

The magnitude of the force F_{2} that q and -q, separated by a distance 2 d, exert on each other is given by Coulomb’s law:

F_{2} = k (q) (- q) / (2 d)^{2} = – k q^{2} / 4 d^{2} = F / 4 = – 2.5 / 4 = – 0.625 N

Problem 4:

A charge of q = – 4.0 × 10^{-6} is placed in an electric field and experiences a force of 5.5 N [E]

a) What is the magnitude and direction of the electric field at the point where charge q is located?

b) If charge q is removed, what is the magnitude and direction of the force exerted on a charge of – 2q at the same location as charge q?

__Solution to Problem 4:__

a) The force on a charge q due to an electric field E is given by

F = q E

The magnitude of E is given by

| E | = | F | / | q | = 5.5 / (4.0 × 10^{-6}) = 1.375 × 10^{6} N / C

Since charge q is negative F and E have opposite direction. E is 1.375 × 10^{6} N / C [W]

b) The force on a charge -2q due to an electric field E is given by

F_{2} = -2 q E = -2(q E) = -2(5.5[E]) = -11 [E] or 11 N [W]

Problem 5:

Three charges are located at the vertices of a right isosceles triangle as shown below. What is the magnitude and direction of the resultant electric field at the midpoint M of AC?

__Solution to Problem 5:__

The magnitude of an electric field due to a charge q is given by.

E = k q / r^{2}

and it is directed away from charge q if q is positive and towards charge q if q is negative. Hence the diagram below showing the direction the fields due to all the three charges. The total field field E is the vector sum of all three fields: E_{AM}, E_{CM} and E_{BM}

_{AM}+ E

_{CM}+ E

_{BM}(vector sum)

Because of the symmetry at point M, E_{AM} and E_{CM} are equal in magnitudes and opposite directions. Hence

E_{AM} + E_{CM} = 0 (vector addition)

Hence

E = E_{BM}

The magnitude of E_{BM} is given by

E_{BM} = k (2 × 10^{-6}) / BM^{2}

Note that

∠MBA = ∠MAB = 45° and MB = MA (lengths)

hence

MB = MA

Pythagora’s theorem

5^{2} = MB^{2} + MA^{2}

MB = 5/√2

MB^{2} = 25/2 = 12.5 cm^{2} = 12.5×10^{-4} m^{2}

The magnitude of the field E_{BM} at point M due to the charge at B is given by

E_{BM} = k q / MB^{2} = k (2 × 10^{-6}) / (12.5 × 10^{-4)} = 9.00 × 10^{9} × 2 × 10^{-6} / (12.5 × 10^{-4}) = 1.44 × 10^{7} N/C

The magnitude of the total field E at M is equal to E_{BM}. Hence

E = E_{BM} = 1.44 × 10^{7} N/C

Problem 6:

What distance must separate two charges of + 5.6×10^{-4}C and -6.3×10^{-4} C in order to have an electric potential energy with a magnitude of 5.0 J in the system of the two charges?

__Solution to Problem 6:__

The magnitude of the electric potiential energy E_{p} of a system of two charges q_{1} and q_{2} separated by a distance r is given by

E_{p} = k | q_{1} | | q_{2} | / r

Solve for r.

r = k q_{1} q_{2} / E_{p} = 9.00×10^{9}×5.6×10^{-4}×6.3×10^{-4} / 5.0 = 6.35×10^{2} m

Problem 7:

The distance between two charges q_{1} = + 2 μC and q_{2} = + 6 μC is 15.0 cm. Calculate the distance from charge q_{1} to the points on the line segment joining the two charges where the electric field is zero.

__Solution to Problem 7:__

At a distance x from q1 the total electric filed is the vector sum of the electric E_{1} from due to q_{1} and directed to the right and the electric field E_{2} due to q_{2} and directed to the left. The vector sum is equal to zero if the magnitudes of the the two fields E_{1} and E_{2} are equal since they have opposite direction.

_{1}and E

_{2}are given by

E_{1} = k q1 / x^{2}

E_{2} = k q2 / (15 – x)^{2}

E_{1} = E_{2} gives the equation

k q1 / x^{2} = k q2 / (15 – x)^{2}

Cross multiply and simplify to obtain

q1(15 – x)^{2} = q2 x^{2}

Which may be written as

(15 – x)^{2} / x^{2} = q2 / q1 = 3

We now to solve for x the equation

(15 – x) / x = ~+mn~ √3

The above equation gives two solutions but only one is positive and is equal to

x = 15 / (1 + √3) ≈ 5.50 cm

Problem 8:

The distance AB between charges Q1 and Q2 shown below is 5.0 m. How much work must be done to move charge Q2 to a new location at point C so that the distance BC = 2.5 m?

__Solution to Problem 8:__

If W is the work to be done to move Q2 from a position where its potential energy is E_{p1} and kinetic energy 0 (from rest) to another position where its potential energy is E_{p2} and kinetic energy 0 (to rest), then by the conservation of energy, we have.

E_{p1} + W = E_{p2}

which gives

W = E_{p2} – E_{p1}

E_{p1} = k Q1 Q2 / AB , with AB = 5 m

E_{p2} = k Q1 Q2 / AC , with AC = 7.5 m

W = k Q1 Q2 (1/AB – 1/AC) = 9.00×10^{9}×5×10^{-6}×-3×10^{-6} (1/7.5 – 1/5) = 9×10^{-3}J

Problem 9:

Two parallel plates separated by distance of 1 cm have a potential difference of 20 V between them. The plates are held in a horizontal position with the negative plate above the positive plate. An electron is released from rest in the upper plate.

a) What is the acceleration of the electron?

b)How long it takes to reach the lower plate?

c)What is the kinetic energy of the electron when it hits the lower plate?

Note: charge of electron q = -1.6×10^{-19}C , mass of electron m = 9.11×10^{-31}Kg

__Solution to Problem 8:__

a) Electric force F exerted on the electron is given by:

F = q E

with the electric field E between the plates given by:

E = ΔV / d = 20 v / 1 cm = 2000 v/m or N/C

F = -1.6×10^{-19}C×2000 N/C = -3.2×10^{-16}N

__Note:__ the weight of the electron is given by: m g = 9.11×10^{-31}×9.8 = 8.93×10^{-30}N

which is much smaller that the electric force acting on the electron and will therefore be neglected.

The acceleration due to the electric force is:

a = F / m = -3.2×10^{-16}/9.11×10^{-31} = – 3.51×10^{14} m/s^{2}

b) Let v_{f} and v_{i} be the final (at lower plate) and initial (from rest upper plate ) velocities of the electron. Hence using the formula

v_{f}^{2} = v_{i}^{2} + 2 a h , h is the distance between the plates.

v_{i} = 0 (from rest)

v_{f}^{2} = 2 a h = 2(- 3.51×10^{14} m/s^{2})(-1×10^{-2}m) (y axis upward)

v_{f} = 2.64×10^{6} m/s

With uniform acceleration a and initial velocity equal to zero, we have the velocity as a function of time given by

v = a t

If T is the time it takes the electron to move from the upper to the lower plates, then

v_{f} = a T

which gives

T = v_{f} / a = 2.64×10^{6} m/s / 3.51×10^{14} m/s^{2} = 7.52×10^{-9}s

c) The kinetic energy E_{k} of the electron at the lower plate is given by

E_{k} = (1/2) m v_{f}^{2} = 0.5×9.11×10^{-31}(2.64×10^{6})^{2} = 3.17×10^{-18}J

Problem 10:

Two electrons are held 3μm apart. When released from rest, what is the velocity of each electron when they are 8μm apart?

__Solution to Problem 10:__

Let E_{p1} be the potential electric energy at rest (distance r = 3μm) and E_{p2} be the potential electric energy when they are 5μm apart and moving. The total (potential and kinetic energies) at each position are given by

E_{t1} = E_{p1} + (1/2) m (0)^{2} = E_{p1}

E_{t2} = E_{p2} + (1/2) m v^{2} + (1/2) m v^{2} = E_{p2} + m v^{2}

Formula for electric potential energy due to charges q1 and q2 distant by r is:

E_{p} = k q1 q2 /r

No external energy is used and no energy is lost, therefore there is conservation of energy such that potential energy is converted into kinetic energy.

E_{p1} = E_{p2} + m v^{2} , v is the velocity when 8μm apart.

charge of electron = – e = -1.60×10^{-19}C , mass of electrom m = 9.109×10^{-31}Kg

m v^{2} = E_{p1} – E_{p2} = k×e×e / (3×10^{-6}) – k×e×e / (5×10^{-6}) = 9×10^{9}(1.6×10^{-19})^{2} [ 1 / (3×10^{-6}) – 1 / (8×10^{-6}) ]

v ≈ 3.48×10^{4} m/s