Physics, asked by lalithakummari5075, 3 months ago

two point charges Q1 and Q2 having masses M and 4m are along a vertical line the magnitude of q2 for the particles to be in vertical equilibrium is​

Answers

Answered by umeshnirmal04
4

Answer:

Let initial separation between the particles be l and this remains constant during the motion. If relative separation remains constant then relative velocity should be zero; hence relative acceleration of the particles should also be zero.

a

r

=a

1

−a

2

=0

or

m

1

q

1

E−F

e

=

m

2

F

e

−q

2

E

or

m

1

m

2

(q

1

m

2

+q

2

m

1

)E

=F

e

[

m

1

1

+

m

2

1

]

or (q

1

m

2

+q

2

m

1

)E=F

e

(m

1

+m

2

)=

4πε

0

1

l

2

q

1

q

2

(m

1

+m

2

)

After solving we get, l=

2

1

πε

0

E(q

2

m

1

+q

1

m

2

)

q

1

q

2

(m

1

+m

2

)

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