Questions: 1. An electrostatic charge of 21 u C is placed at a distance in air of 15 cm



From a second charge.  The force of attraction between the two charges is 26 N.



A.      Calculate the magnitude of the second charge.



Sol) q₁ = 21 u C = 21 x 10^-6 C

q₂ = ?, F = 26N, R = 15cm = 0.15m

Using F = q₁ q₂ / 4πε_oR²

1/4πε_o = 9 x 10⁹ N-m²-C^-2

Thus F = 9 x 10⁹ q₁ q₂ / ( 0.15)²

or 26 = 9 x 10⁹ q₁ q₂ / ( 0.15)²

or 26 = 9 x 10⁹ x 21 x 10^-6  q₂ / ( 0.15)² = 1.89 x 10⁵ q₂ / ( 0.15)²

or q₂ = 26 x ( 0.15)² / 1.89 x 10⁵

or q₂ = 3.1 x 10^-6 C = 3.1 μC

B       If the distance between the charges is decreased to 5.0 cm, calculate the magnitude of the new force acting on each ball.

Sol) Force between the balls when distance is 15cm = F₁ = 26N, R₁ = 15cm

Let force between the balls when distance is 5cm be = F₂ , R₂ = 5cm

Using relation, F = q₁ q₂ / 4πε_oR²

we have F₁ = q₁ q₂ / 4πε_oR₁ ²

and F₂ = q₁ q₂ / 4πε_oR₂ ²

Thus F₁ / F₂ = R₂ ² / R₁ ²

or 26 / F₂ = 5² / 15² = 25 / 225 = 1/9

or F₂ = 26*9 = 234 N

Now

2. An electron, moving through an electric field, experiences an acceleration of 6.3 x 10³ m/s².

A. How much electrostatic force is acting on the electron?

Sol) Acceleration a = 6.3 x 10³ m/s².

Charge of electron = e = 1.6 x 10^-19 C

Mass of electron = 9.1 x 10^-31 Kg

Let Electric force be = F

Now F = ma

or F = 9.1 x 10^-31 x  6.3 x 10³ = 5.73 x 10^-27 N





B. What is the strength of the electric field?

Sol) Let strength of electric field be E

Now F = e E

but F = 5.73 x 10^-27 N  (as calculated above in part A)

or E = F/e = 5.73 x 10^-27 / 1.6 x 10^-19

or E = 3.58 x 10^-8 N/C

4.      A 3.75 x 10^{- 9} F capacitor carries a charge of 1.75x10 ^–8 C

       A.      What is the potential difference across the plates?

Sol) C = 3.75 x 10^{- 9} F

Q = 1.75x10 ^–8 C

Now Using Q = CV

we have V = Q/C = 1.75x10 ^–8 / 3.75 x 10^{- 9} = 4.67 V

B.       If the plates are 6.50 x 10^-4 m a part, what is the strength of the



 Electric field between them?

Sol) Distance between the plates, d = 6.50 x 10^-4 m

V = 4.67 V (as calculated above in part A)

Thus E = V/d = 4.67 / 6.50 x 10^-4 = 7.18 x 10³ V/m







5.      A charge of –3.00 u C is fixed at the centre of a compass.  Two



Additional charges are fixed on the circle of the compass (radius = 0.100m).  The charges on the circle are –4.00 u C at the position due north and + 5.00 u C at a position due east.  What is the magnitude and direction of the net electrostatic force acting on the charge at the centre?

Sol) q = –3.00 u C = -3 x 10^-6 C

Bsdsd

           

A

       C(q)

Here BC = AC = 0.1m

Here charge q is placed at C (shown in above figure)

Let at point B which is north to C charge put be = q₁ = -4 uC = -4 x 10^-6 C

also at point A which is east to C charge kept = q₂ = 5 x 10^-6 C

Forces at charge q placed at point C due to charges at B and A are mutually perpendicular to each other.

F_B = 9 x 10⁹ q.q₁ / R²

or F_B = 9 x 10⁹ x (-3 x 10^-6) x (-4 x 10^-6) / (0.1)²

or F_B = 10.8 N long BC

Similarly force on q due to q₂ at B = F_A = 9 x 10⁹ q.q₂ / R²

or F_A = 9 x 10⁹ x (-3 x 10^-6) x (5 x 10^-6) / (0.1)²

or F_A = 13.5 N along CA

Thus total force F on charge q placed at C is given by

F =

or F =

or F = 17.28 N

The direction of the force is such that it makes an angle θ with CA given by

θ = tan^-1 F_A/F_B = tan^-1 13.5/10.8 = tan^-1 1.25

or θ = 51.34 degrees.

Hence the direction of the force F = 17.28 N is such that it makes an angle 51.34 deg with CA

6. A particle (mass = 2.5 x 10^-15 kg) with a negative charge of 3.5 x 10^-5 C is moved from the positive plate of a parallel plate capacitor to the negative plate, a distance of 2.5 cm.



A. If the potential difference between the two plates is 15 volts, how much work is done in moving the particle?

Sol) m = 2.5 x 10^-15 kg

q = -3.5 x 10^-5 C

Distance between the capacitor plates, d = 2.5 cm = 0.025 m

Potential difference between the plates, V = 15 volts

Thus intensity of electric field E = V/d = 15/0.025 = 600 V/m

Work done in moving the particle = W = F.d  = qE.d

or W = qE.d = 3.5 x 10^-5 x 600 x 0.025 = 5.25 x 10^-4 J is the amount of work done in moving the particle.

B. If the particle is released from the negative plate, it will accelerate toward the positive plate.  What will be the velocity of the particle just before it strikes the positive plate?

Sol) m = 2.5 x 10^-15 kg

q = -3.5 x 10^-5 C

V = 15 volts

E = 600V/m (as calculated above)

Now acceleration of the charged particle = a = qE/m

or a = 3.5 x 10^-5 x 600 / 2.5 x 10^-15

or a = 8.4 x 10¹² m/s²

Using the relation v² – u² = 2aS

here S = d = 0.025 m

u = 0

Thus v² – 0 = 2 x 8.4 x 10¹² x 0.025

or v² = 4.2 x 10¹¹ m/s

or v = 6.48 x 10⁵ m/s



C. What is the strength of the electric field between the plates?

Sol) Electric field between the plates E = V/d = 15/0.025

or E = 600 V/m

D. What force will be applied to the particle as it accelerates toward the Negative plate?

Sol) Force on the particle F is given by

F = qE

Here, q = -3.5 x 10^-5 C

E = 600 V/m

Thus F = 3.5 x 10^-5 x 600 N = 2.1 x 10^-2 N = 0.021 N