KCET · Physics · Electrostatics
A \(2 \mathrm{~g}\) object, located in a region of uniform electric field \(\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}\) carries a charge \(Q\). The object released from rest at \(x=0\), has a kinetic energy of \(0.12 \mathrm{~J}\) at \(x=0.5 \mathrm{~m}\). Then, \(Q\) is
- A \(400 \mu \mathrm{C}\)
- B \(-400 \mu \mathrm{C}\)
- C \(800 \mu \mathrm{C}\)
- D \(-800 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(800 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Given, \(m=2 g=2 \times 10^{-3} \mathrm{~kg}\)
\(\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}\)
At, \(x=0,(\mathrm{KE})_{1}=0\) and \(x=0.5 \mathrm{~m},(\mathrm{KE})_{2}=0.12 \mathrm{~J}\)
From work-energy theorem,
work done \(=\) change in kinetic energy
\(\Rightarrow \text {Force } \times \text { displacement }=\Delta \mathrm{KE} \)
\( Q E \times d =\mathrm{KE}_{2}-\mathrm{KE}_{1} \)
\( \Rightarrow Q \times 300 \times 0.5=0.12 \)
\( \text {or } Q=\frac{0.12}{300 \times 0.5}=800 \mu \mathrm{C}\)
The charge \(Q\) is moving in the direction of electric field. Hence, it is positive.
\(\mathbf{E}=\left(300 \mathrm{NC}^{-1}\right) \hat{\mathbf{i}}\)
At, \(x=0,(\mathrm{KE})_{1}=0\) and \(x=0.5 \mathrm{~m},(\mathrm{KE})_{2}=0.12 \mathrm{~J}\)
From work-energy theorem,
work done \(=\) change in kinetic energy
\(\Rightarrow \text {Force } \times \text { displacement }=\Delta \mathrm{KE} \)
\( Q E \times d =\mathrm{KE}_{2}-\mathrm{KE}_{1} \)
\( \Rightarrow Q \times 300 \times 0.5=0.12 \)
\( \text {or } Q=\frac{0.12}{300 \times 0.5}=800 \mu \mathrm{C}\)
The charge \(Q\) is moving in the direction of electric field. Hence, it is positive.
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