A $0.50 g$ ball carries a charge of $+ 10 μ C$ . It is suspended by a string in a downward electric field of intensity $300 {NC}^{- 1}$ . Then the tension in the string is $x \times 10^{- 3} N$ . The value of $x$ is _____.

$(g = 10 {ms}^{- 2})$

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Solution

From the FBD of the charge, we get,

Tension in the string $=$ Weight + Force due to field

$\Rightarrow T = m g + q E$

$= 0.50 \times 10^{- 3} \times 10 + 10 \times 10^{- 6} \times 300$

$= 8 \times 10^{- 3} N$

$∴ x = 8$

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A 0.50 g ball carries a charge of +10 μC. It is suspended by a string in a downward electric field of intensity 300 NC−1. Then the tension in the string is x×10−3 N. The value of x is _____. (g=10 ms−2)

From the FBD of the charge, we get, Tension in the string = Weight + Force due to field ⇒T=mg+qE =0.50×10−3×10+10×10−6×300 =8×10−3 N ∴ x=8

A $0.50 g$ ball carries a charge of $+ 10 μ C$ . It is suspended by a string in a downward electric field of intensity $300 {NC}^{- 1}$ . Then the tension in the string is $x \times 10^{- 3} N$ . The value of $x$ is _____.

$(g = 10 {ms}^{- 2})$

From the FBD of the charge, we get,

Tension in the string $=$ Weight + Force due to field

$\Rightarrow T = m g + q E$

$= 0.50 \times 10^{- 3} \times 10 + 10 \times 10^{- 6} \times 300$

$= 8 \times 10^{- 3} N$

$∴ x = 8$