MHT CET · Physics · Electrostatics
A particle ' \(A\) ' has charge ' \(+q\) ' and a particle ' \(B\) '. has charge ' \(+4 q\) '. Each has same mass ' m '. When they are allowed to fall from rest through the same potential, the ratio of their speeds will become (particle A to particle B)
- A \(2: 1\)
- B \(1: 2\)
- C \(1: 4\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(B) \(1: 2\)
Step-by-step Solution
Detailed explanation
Electric force on charged particle placed in a electric field.
\(F=q E \)
\( F_A=+q E \)
\( F_A=m_A a_A \)
\( \therefore a_A =\frac{q E}{m} \)
\( F_B =+4 q E \)
\( F_B =m_B a_B \)
\(a_B=\frac{4 q E}{m} \)
\( \therefore a_A=\frac{a_B}{4}...(i)\)
As both the bodies are falling from rest and cover same distance.
\(V_A^2=0+2 a_A x\)
\(V_B^2=0+2 a_B x\)
\(\therefore\left(\frac{V_A}{V_B}\right)^2=\frac{a_A}{a_B}\)
\(\left(\frac{V_A}{V_B}\right)^2=\frac{a_B}{4 \times a_B}\)
\(\therefore\frac{V_A}{V_B}=\frac{1}{2}\)
...[From (i)]
\(F=q E \)
\( F_A=+q E \)
\( F_A=m_A a_A \)
\( \therefore a_A =\frac{q E}{m} \)
\( F_B =+4 q E \)
\( F_B =m_B a_B \)
\(a_B=\frac{4 q E}{m} \)
\( \therefore a_A=\frac{a_B}{4}...(i)\)
As both the bodies are falling from rest and cover same distance.
\(V_A^2=0+2 a_A x\)
\(V_B^2=0+2 a_B x\)
\(\therefore\left(\frac{V_A}{V_B}\right)^2=\frac{a_A}{a_B}\)
\(\left(\frac{V_A}{V_B}\right)^2=\frac{a_B}{4 \times a_B}\)
\(\therefore\frac{V_A}{V_B}=\frac{1}{2}\)
...[From (i)]
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