Home
Class 12
PHYSICS
In free space, a particle A of charge 1 ...

In free space, a particle A of charge `1 mu C`, is held fixed at a point P. Another particle B of the same charge and mass `4 mu g`,is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is: `[take 1/(4pi epsilon_0)=9xx10^9 Nm^2 C^(-2)]`

A

`1. 5 xx 10^(2) m//s `

B

`1.0 m //s`

C

`3.0 xx 10^(4) m//s`

D

`2.0 xx10^(3) m//s`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

A particle A of charge 1 mu C is held fixed at a point P in free space . Another particle B of same charge and mass 4 mu g is kept at a distance of1mm from P .If B is released then its velocity at a distance of 9 mm from P is (Take (1)/( 4pi epsilon_(0)) =9 xx 10^(9)Nm^(2) C^(-2))

Two charge particle P & Q having same charge 1muC and mass 4mukg are initially kept at the distance of 1mm . Charge P is fixed, then the velocity of charge parti Q when the separationbetween then becmoes 9mm .

The electric field due to a charge at a distance of 3 m from it is 500 NC^(-1) . The magnitude of the charge is [(1)/(4 pi epsi_(0)) = 9 xx 10^(9) N - m^(2)//C^(2)]

A hollow sphere of radius 0.1 m has a charge of 5 xx10^(-8) C . The potential at a distance of 5 cm from the centre of the sphere is (1/(4pi epsi_(0))=9xx10^(9) Nm^(2) C^(-2))

A particle P of charge q and m , is placed at a point in gravity free to move. Another particle Q, of same charge and mass , is projected from a distance r from P with an initial speed v_(0) towards P , initial the distance between P and Q decreases and then increases. The potential energy of the system of particles P and Q, at closest separation is

A particle of mass 2 g and charge 1 muC is held at rest on a frictionless surface at a distance of 1m from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at distance of 10 m from fixed charge is :

What is the magnitude of a point charge due to which the electric field 30cm away the magnitude 2 ? [1//4 pi epsilon_(0)=9xx10^(9)Nm^(2)//C^(2)]

Three particles, each having a charge of 10 mu C are placed at the coners of an equilateral triangle of side 10 cm . The electrostatic potential energy of the system is (Given (1)/(4pi epsilon_(0)) = 9 xx 10^(9)N-m^(2)//C^(2))

Two identical charges, 5 muC each are fixed at a distance 8 cm and a charged particle of mass 9 xx 10^(-6) kg and charge -10 muC is placed at a distance 5 cm from each of them and is released. Find the speed of the particle when it is nearest to the two charges.

In the rectangle, shown below, the two corners have charges q_(1) = - 5mu C and q_(2) = + 2.0 mu C . The work done in moving a charge +3.0 mu C from B to A is (take 1//4 4pi epsilon_(0) = 10^(10) N-m^(2)//C^(2))