The distance between charges $5 \times 10^{- 11} C$ is 0.2m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is (the distance from smaller charge)

0.4 m towards the larger charge

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.6 m on the other side

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 m on the other side

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.3 m to wards larger charge

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.5 m on the other side
$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} - 1} = \frac{1.2}{\sqrt{\frac{5 \times 10^{- 11}}{2.5 \times 10^{- 11}}} - 1} = \frac{0.2}{\sqrt{2} - 1} = \frac{0.2}{0.414}$
x = 0.5m on the other side.

Suggest Corrections

Similar questions

Q. The distance between charges

5 \times 10^{- 11}

C and

- 2.7 \times 10^{- 11}

C is

0.2 m

. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

Q. The distance between charges

5 \times 10^{- 11}

C and

- 2.7 \times 10^{- 11}

C is

0.2 m

. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

Two charges $2 μ C$ and $1 μ C$ are placed at a distance of 10cm. The distance of third charge from one of the charges so that it does not experience any force is:

Q. Two positive and equal charges are fixed at certain distance. A third charge (small) is placed in between the line joining of the two charges and it experiences zero net force due to the other two.

Q. Two charges of

40 μ C

and

90 μ C

are placed at a distance of

1 m

from each other. Where should a third charge be placed on the line joining them, if total force on the third charge is zero?

Join BYJU'S Learning Program

The distance between charges 5×10−11C is 0.2m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is (the distance from smaller charge)

The correct option is C 0.5 m on the other sidex=d√q2q1−1=1.2√5×10−112.5×10−11−1=0.2√2−1=0.20.414 x = 0.5m on the other side.

The distance between charges $5 \times 10^{- 11} C$ is 0.2m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is (the distance from smaller charge)

The correct option is C 0.5 m on the other side
$x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} - 1} = \frac{1.2}{\sqrt{\frac{5 \times 10^{- 11}}{2.5 \times 10^{- 11}}} - 1} = \frac{0.2}{\sqrt{2} - 1} = \frac{0.2}{0.414}$
x = 0.5m on the other side.