KCET · Physics · Electrostatics
The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at \(A\) is \(40 \mathrm{Vm}^{-1}\), what is the force on \(20 \mu \mathrm{C}\) charge kept at \(B\) ?
- A \(4 \times 10^{-4} \mathrm{~N}\)
- B \(8 \times 10^{-4} \mathrm{~N}\)
- C \(16 \times 10^{-4} \mathrm{~N}\)
- D \(1 \times 10^{-4} \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(4 \times 10^{-4} \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
According to given figure,
Electric field at point \(A, E_{A}=40 \mathrm{Vm}^{-1}\)
Since, electric field lines on the left have twice the separation on those on the right (at point \(B\) ), hence electric field at point \(B\)
\(\begin{aligned}
&E_{B}=\frac{E_{A}}{2}=\frac{40}{2} \\
&E_{B}=20 \mathrm{Vm}^{-1}
\end{aligned}\)
Force on charge \(q\) kept at \(B\) is
\(F=q E_{B} \)
\( \text {Given, } q=20 \mu \mathrm{C}=20 \times 10^{-6} \mathrm{C} \)
\( \Rightarrow F=20 \times 10^{-6} \times 20=4 \times 10^{-4} \mathrm{~N}\)
Electric field at point \(A, E_{A}=40 \mathrm{Vm}^{-1}\)
Since, electric field lines on the left have twice the separation on those on the right (at point \(B\) ), hence electric field at point \(B\)
\(\begin{aligned}
&E_{B}=\frac{E_{A}}{2}=\frac{40}{2} \\
&E_{B}=20 \mathrm{Vm}^{-1}
\end{aligned}\)
Force on charge \(q\) kept at \(B\) is
\(F=q E_{B} \)
\( \text {Given, } q=20 \mu \mathrm{C}=20 \times 10^{-6} \mathrm{C} \)
\( \Rightarrow F=20 \times 10^{-6} \times 20=4 \times 10^{-4} \mathrm{~N}\)
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