Question

calculate the pd to be maintained between two horizontal conducting plates separated by distance of 10mm so that small charged oil drop of mass 1.31×10-14 kg will remain in equilibrium. charge in equilibrium is 6.4×10-19

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Answer 1

Answer:

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Explanation:

Solution

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Correct option is

C

20 ms

−2

Force in electric field is given by,

F=

qE

and

F=

mg

(By gravitation)

So,mg=

qE

(for equilibrium)

2.4×10

−18

4.8×10

−10

×10

−3

×10

=

E

⇒ E= 2× 10

6

m

V

Now when polarity is changed force from electric field also comes in the direction of mg.

So,

F

net

=

mg+

qE=

ma

(By Newton's Law)

⇒

g+

m

q

E=

a

a=

10+

4.8×10

−10

×10

−3

2.4×10

−18

×2×10

6

=

10+

10

=

20m/s

2

Answer 2

Answer:

Explanation: i need answers