A thin glass rod is bent in a semicircle of radius R. A charge is nonuniformly distributed along the rod with a linear charge density \lambda=\lambda_0 \sin \theta ( \lambda_0 is a positive constant). The electric field at the centre P of the semicircle is,

Correct Answer is : −λo8πεoR^j

-\frac{\lambda_o}{8 \pi \epsilon_o R} \stackrel{\^}{j}

\frac{\lambda_o}{8 \pi \epsilon_0 R} \stackrel{\^}{j}

\frac{\lambda_0}{8 \pi \epsilon_0 {\R}} \stackrel{\^}{ {\i}}

-\frac{\lambda_0}{8 \pi \epsilon_0 R} \stackrel{\^}{ {\i}}

WBJEE 2023 Physics & Chemistry Solved Paper

Question : 61 of 80

Marks: +1, -0

A thin glass rod is bent in a semicircle of radius R. A charge is nonuniformly distributed along the rod with a linear charge density λ=λ0sin‌θ ( λ0 is a positive constant). The electric field at the centre P of the semicircle is,

−‌
λo
8πεoR
^
j
‌
λo
8πε0R
^
j
‌
λ0
8πε0R
^
i
−‌
λ0
8πε0R
^
i

Validate

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