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

Byju's Answer

Standard XII

Physics

Uniform Circular Motion

A small parti...

Question

A small particle has charge -5.00 $μ$ C and mass $2.00 \times 10^{4}$ kg. It moves from point A, where the electric potential is $V_{A} = + 200 V$ ,to point B, where the electric potential is $V_{B} = + 800 V .$ The electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

Open in App

Solution

by conservation of energy
$\frac{1}{2} m v_{A}^{2} + q V_{A} = \frac{1}{2} m v_{B}^{2} + q V_{B}$
$\frac{1}{2} m (v_{B}^{2} - v_{A}^{2}) = q (V_{A} - V_{B}) = + .003 J$
Therefore , Speed at B is faster.
Solving above
$v_{B} = 7.42 m / s$

Suggest Corrections

Similar questions

Q. The electric potential at a certain distance from a point charge

Q

810 V

and electric field is

300 N / C

. The minimum speed with which a particle of charge

- Q

and mass

m = 6 \times 10^{- 16} k g

should be projected from that point so that it moves into the field free region is:

Q. In figure, a particle has mass

m = 5 g

and charge

q^{'} = 2 \times 10^{- 9} C

starts from rest at point

a

and moves in a straight line to point

b

. What is its speed

v

at point

b

Q. A particle

A

has charge

+ q

and particle

B

has charge

+ 4 q

with each of them having the same mass

m

. When allowed to move from rest through same electrical potential difference, the ratio of their speeds

\frac{V_{A}}{V_{B}}

will become

Q. The kinetic energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle.

The kineti energy of a charged particle decreases by 10 J as it moves from a point at potential 100 V to a point at potential 200 V. Find the charge on the particle.

Join BYJU'S Learning Program

by conservation of energy 12mv2A+qVA=12mv2B+qVB12m(v2B−v2A)=q(VA−VB)=+.003JTherefore , Speed at B is faster.Solving above vB=7.42m/s

by conservation of energy
$\frac{1}{2} m v_{A}^{2} + q V_{A} = \frac{1}{2} m v_{B}^{2} + q V_{B}$
$\frac{1}{2} m (v_{B}^{2} - v_{A}^{2}) = q (V_{A} - V_{B}) = + .003 J$
Therefore , Speed at B is faster.
Solving above
$v_{B} = 7.42 m / s$