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Relation between Work Done and PE

A pendulum bo...

Question

A pendulum bob of mass $30 m g$ carrying a charge of $2.0 \times 10^{- 8}$ is at rest in a uniform horizontal electric field of $20000 N / C$ . The tension in the thread of the pendulum is $(g = 10 m / s^{2})$ :

5 \times 10^{- 4} N

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4 \times 10^{- 4} N

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3 \times 10^{- 4} N

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2 \times 10^{- 4} N

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Solution

The correct option is C $5 \times 10^{- 4} N$
$H o r i z o n t a l f o r c e a c t i n g o n C h a r g e = q E = 2 \times 10^{- 8} \times 20000$
$F_{H} = 4 \times 10^{- 4} N$

$V e r t i c a l f o r c e a c t i n g o n c h a r g e = m g = 30 \times 10^{- 6} \times 10 = 3 \times 10^{- 4} N$

Tension will be equal to resultant of electrostatic force and force due to gravity,

$F_{r e s} = \sqrt{q E^{2} + m g^{2}} = 5 \times 10^{- 4} N$

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A pendulum bob of mass 30mg carrying a charge of 2.0×10−8 is at rest in a uniform horizontal electric field of 20000N/C. The tension in the thread of the pendulum is (g=10m/s2):

The correct option is C 5×10−4NHorizontal force acting on Charge=qE=2×10−8×20000 FH=4×10−4NVertical force acting on charge=mg=30×10−6×10=3×10−4NTension will be equal to resultant of electrostatic force and force due to gravity, Fres=√qE2+mg2=5×10−4N

A pendulum bob of mass $30 m g$ carrying a charge of $2.0 \times 10^{- 8}$ is at rest in a uniform horizontal electric field of $20000 N / C$ . The tension in the thread of the pendulum is $(g = 10 m / s^{2})$ :