You visited us 1 times! Enjoying our articles? Unlock Full Access!

Principle of Superposition

Three charges...

Question

Three charges $q_{1} = 3 m C, q_{2} = - 3 m C$ and $q_{3}$ are kept at the vertices of a triangle as shown in the figure. lf the net force acting on $q_{1}$ is $¯ ¯¯ ¯ F$ , the charge $q_{3}$ would have the magnitude $($ 1 + $\frac{1}{n})^{2} m C$ . So, $n$ is:

8

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

4

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

1

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

16

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

Open in App

Solution

The correct option is A $8$
Let $_{1}$ be the force on $q_{1}$ due to $q_{2}$ and $_{2}$ be that due to $q_{3}$
As, $q_{1} > 0, q_{2} < 0,$ the charge $q_{3}$ should be + ve to get the net force $\to F$ .
$F_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{4^{2}}, F_{2} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{3}}{3^{2}}$
From the figure, $tan θ = \frac{F_{2}}{F_{1}}$
$∴ \frac{3}{4} = \frac{q_{3}}{9} \times \frac{16}{q_{2}}$
$∴ q_{3} = \frac{27}{64} \times q_{2} = \frac{81}{64} = {(\frac{9}{8})}^{2} m C$
$= {(1 + \frac{1}{8})}^{2} m C$
Hence, $n = 8$

Suggest Corrections

Similar questions

Q. Three charges

- q_{1}, + q_{2}

and

- q_{3}

are placed as shown in the figure. The x-component of the force on

- q_{1}

is proportional to

Q. Three point charges

q_{1}, q_{2}

and

q_{3}

are taken such that when

q_{1}

and

q_{2}

are placed close together to form a single point charge, the force on

q_{3}

at distance

L

from this combination is a repulsion of

2

unit magnitude. When

q_{2}

and

q_{3}

are combined, the force on

q_{1}

, from same distance

L

is of magnitude

4

unit and is attractive. Also

q_{3}

and

q_{1}

combining exert

6

a force of magnitude

18

unit attractive on

q_{2}

from same distance

L

, Algebraic ratio of charges

q_{1}, q_{2}

and

q_{3}

is:

Q. Three point charges

- q_{1}

+ q_{2}

and

- q_{3}

are placed as shown in the figure. The x-component of force on

- q_{1}

is proportional to

Q. Three charges

- q_{1}, + q_{2}, - q_{3}

are placed as shown in the figure. The x-component on the force on

- q_{1}

is proportional to:

Three point charges $Q_{1}$ , $Q_{2}$ and $Q_{3}$ in that order are placed equally spaced along a straight line. $Q_{2}$ and $Q_{3}$ are equal in magnitude but opposite in sign. If the net force on $Q_{3}$ is zero, the value of $Q_{1}$ is :

Join BYJU'S Learning Program

Three charges q1=3 mC,q2=−3 mC and q3 are kept at the vertices of a triangle as shown in the figure. lf the net force acting on q1 is ¯¯¯¯F, the charge q3 would have the magnitude (1 +1n)2mC. So, n is:

The correct option is A 8Let →F1 be the force on q1 due to q2 and →F2 be that due to q3As, q1>0,q2<0, the charge q3 should be + ve to get the net force →F.F1=14πε0q1q242,F2=14πε0q1q332From the figure, tanθ=F2F1∴34=q39×16q2∴q3=2764×q2=8164=(98)2mC=(1+18)2mCHence, n=8

Three charges $q_{1} = 3 m C, q_{2} = - 3 m C$ and $q_{3}$ are kept at the vertices of a triangle as shown in the figure. lf the net force acting on $q_{1}$ is $¯ ¯¯ ¯ F$ , the charge $q_{3}$ would have the magnitude $($ 1 + $\frac{1}{n})^{2} m C$ . So, $n$ is: