Ten positively charged particles are kept fixed on the x-axis at points x = 10 cm , 20 cm , 30 cm . . . . . 100 cm . The first particle has a charge 1 .0 × 10 − 8 C , the second 8 × 10 − 8 C , third 27 × 10 − 8 C , and so on. The tenth particle has a charge 1000 × 10 − 8 C . Find the magnitude of the electric force acting on a 1 C charge placed at the origin.

Given that the charges are increased as, 1 3 × 10 - 8 C , 2 3 × 10 - 8 C , 3 3 × 10 - 8 C . . . . . . . . . . . . . . . . . . . . . 10 3 × 10 - 8 C Now, the net force on 1 C by all the charges on order is in the same direction. Hence, all forces on the given charges will be added to give the net amount of force on the charge. And the net force due to all the charges can be given by the vector sum of all the individual forces by the superposition principle. Using Coulomb's law for finding the interactive force between two charges let, q 1 , q 2 separated by a distance of r is given by, F = K q 1 q 2 r 2 Where, K = 9 × 10 9 N m 2 C - 2 is electrostatic constant Using the above expression for calculation of force between 1 C and 1 × 10 - 8 C Force, F 2 = K × 1 × 1 × 10 - 8 0 . 1 2 Similarly, the interactive force between 1 C and 8 × 10 - 8 C is, F 2 = K × 1 × 2 × 10 - 8 0 . 2 2 Similarly, F 3 = K × 1 × 27 × 10 - 8 0 . 3 2 To calculate the net force on the charge, F = F 1 + F 2 + F 3 + . . . . . . . . . . . + F 10 = K 1 × 10 - 8 0 . 1 2 + 8 × 10 - 8 0 . 2 2 + 27 × 10 - 8 0 . 3 2 + . . . . . . . . . . . . . . . . . . . + 1000 × 10 - 8 1 2 = K × 10 - 8 1 3 0 . 1 2 + 2 3 0 . 2 2 + 3 3 0 . 3 2 + . . . . . . . . . . . . . . . . . . . . . + 10 3 1 2 = K × 10 - 6 1 + 2 + 3 + . . . . . . . . . . . . . . . . . + 10 = K × 10 - 6 × 10 × 11 2 = 55 K × 10 - 6 = 55 × 9 × 10 9 × 10 - 6 = 4 . 95 × 10 5 N

If an oil drop of weight 3 .2 × 10 − 13 N is balanced in an electric field of 5 × 10 5 Vm − 1 find the charge on the oil drop.

Given, Weight of the oil drop W = 3 .2 × 10 − 13 N Electric field strength E = 5 × 10 5 Vm − 1 The weight of the oil drop is balanced by the force due to electric field, ⇒ W = qE ⇒ q = W E = 3 .2 × 10 − 13 N 5 × 10 5 Vm − 1 = 0 .64 × 10 − 18 C So, the charge on the oil drop = 0 .64 × 10 − 18 C

EASY

12th CBSE

IMPORTANT

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Two pith-balls each weighing $10^{- 3} kg$ are suspended from the same point by means of silk threads $0.5 m$ long. On charging the balls equally, they are found to repel each other to a distance of $0.2 m$ . Calculate the charge on each ball.

Important Questions on Electric Charges and Field

MEDIUM

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 1

Ten positively charged particles are kept fixed on the x-axis at points

x = 10 cm, 20 cm, 30 cm . . . . . 100 cm

. The first particle has a charge

1 .0 \times 10^{- 8} C

, the second

8 \times 10^{- 8} C

, third

27 \times 10^{- 8} C

, and so on. The tenth particle has a charge

1000 \times 10^{- 8} C

. Find the magnitude of the electric force acting on a

1 C

charge placed at the origin.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 2

Charges $q_{1} = 1.5 mC, q_{2} = 0.2 mC$ and $q_{3} = - 0 .5 mC$ are placed at the points $A$ , $B$ and $C$ respectively, as shown in Figure. If $r_{1} = 1 .2 m$ and $r_{2} = 0 .6 m$ calculate the magnitude of resultant force on $q_{2}$ .

Question Image

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 3

Two equal positive charges, each of $2 μ C$ interact with a third positive charge of $3 μ C$ situated as shown in Fig. 1.32. Find the magnitude and direction of the force experienced by the charge of $3 μ C$ .

Question Image

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 4

Four charges

+ q, + q, - q

and

- q

are placed respectively at the four corners

A, B, C

and

D

of a square of side

a

. Calculate the force on a charge

Q

placed at the centre of the square.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 1

If an oil drop of weight

3 .2 \times 10^{- 13} N

is balanced in an electric field of

5 \times 10^{5} {Vm}^{- 1}

find the charge on the oil drop.

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 2

Calculate the magnitude of the electric field, which can just balance a deuteron of mass

3 .2 \times 10^{- 27} kg

. Take

g = 10 {ms}^{- 2}

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 3

A charged oil drop remains stationary when situated between two parallel plates

20 mm

apart and a p.d. of

500 V

is applied to the plates. Find the charge on the drop if it has a mass of

2 \times 10^{- 4} kg

. Take

g = 10 {ms}^{- 2}

EASY

12th CBSE

IMPORTANT

New Simplified Physics (Vol 1) For Class 12>Chapter 1 - Electric Charges and Field>Problems For Practice>Q 4

In Millikan's experiment, an oil drop of radius

10^{- 4} cm

remains suspended between the plates which are

1 cm

apart. If the drop has charge of

5 e

over it, calculate the potential difference between the plates. The density of oil may be taken as

1 .5 {gcm}^{- 3}

Two pith-balls each weighing 10−3kg are suspended from the same point by means of silk threads 0.5 m long. On charging the balls equally, they are found to repel each other to a distance of 0.2 m. Calculate the charge on each ball.

Important Questions on Electric Charges and Field

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Two pith-balls each weighing $10^{- 3} kg$ are suspended from the same point by means of silk threads $0.5 m$ long. On charging the balls equally, they are found to repel each other to a distance of $0.2 m$ . Calculate the charge on each ball.