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

0.44 m

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

0.65 m

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

0.556 m

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

0.350 m

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

Open in App

Solution

The correct option is C 0.556 m
If two opposite charges are separated by a certain distance, then for it's equalibrium a third
charge should be kept outside and near the charge which is smaller in magnitude.
Here, suppose third charge $q$ is placed at a distance $x$ from $- 2.7 \times 10^{- 11} C$ then for it's
equalibrium $| F_{1} | = | F_{2} |$

$\Rightarrow \frac{K Q_{1} q}{(x + 0.2)^{2}} = \frac{K Q_{2} q}{x^{2}} \Rightarrow x = 0.556 m$

$(H e r e K = \frac{1}{4 π ϵ_{0}} a n d Q_{1} = 5 \times 10^{- 11} C, Q_{2} = - 2.7 \times 10^{- 11} C)$

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

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)

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 point charges

q_{1} = + 9 μ C

and

q_{2} = - 1 μ C

are held

10 cm

apart. Where should at third charge

+ Q

be placed from

q_{2}

on the line joining them so that charge

Q

does not experience any net force ?

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.

Join BYJU'S Learning Program

The distance between charges 5×10−11C and −2.7×10−11C 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 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