Q. 495.0( 3 Votes )

# A ball of mass 100g and having a charge of 4.9 × 10^{–5} C is released from rest in a region where a horizontal electric field of 2.0 × 10^{4} C NC^{–1} exists.

(a) Find the resultant force acing on the ball.

(b) What will be the path of the ball?

(c) Where will the ball be at the end of 2s?

Answer :

**Given:**Mass of the ball : m = 100 g = 100× 10

^{-3}kg =0.1 kg

Charge on the ball : q = 4.9 × 10

^{–5}C

Horizontal electric field : E= 2.0 × 10

^{4}C NC

^{–1}

Initial Velocity of the ball: u = 0

Here, R is the resultant Force due to gravitational force F

_{g}and electric force F

_{e}.

**Formula used:**

(a)

Electric force F

_{e}due to charge q and electric field E is

F

_{e}= qE

Gravitational force F

_{g}experience due to mass of the ball m and acceleration due to gravity g is:

F

_{g}=mg

Thus we see that F

_{e}= F

_{g}

The Resultant force R can be calculated by:

R

^{2}=F

_{e}

^{2}+F

_{g}

^{2}

∴ R^{2}= (0.98)^{2}+(0.98)^{2}

∴R=√1.9208

∴R=1.3859 N

Hence the resultant force of 1.3859 N is acting on the ball.

(b)

We take tangent of the angle θ

But F_{g}= F_{e},

Hence, the path of the ball is along a straight line and inclined at an angle of 45° with the horizontal electric field.

(c)

Here, we will be using one of the equations of motion.

Here s is the distance covered by the ball, u is the initial velocity of the ball, a is the acceleration of the ball and t is the time required to cover s.

We need to find s at t=2s

Firstly ,vertical displacement due to gravitational force is:a=g

Secondly,

Horizontal displacement due to electric force is:

Thus net displacement = (s_{v}^{2}+ s_{h}^{2})^{1/2}

∴ Net displacement = ((19.6)^{2}+(19.6^{2}))^{1/2}

∴ Net displacement = 27.71 m

Thus, the ball will be at a distance of 27.71m after 2s

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The electric force experienced by a charge of 1.0 × 10^{–6} C is 1.5 × 10^{–3} N. Find the magnitude of the electric field at the position of the charge.

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HC Verma - Concepts of Physics Part 2