JEE MAIN - Physics (2006 - No. 34)
Two insulating plates are both uniformly charged in such a way that the potential difference between them is $${V_2} - {V_1} = 20\,V.$$ (i.e., plate $$2$$ is at a higher potential). The plates are separated by $$d=0.1$$ $$m$$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate $$1.$$ What is its speed when it hits plate $$2$$?
$$\left( {e = 1.6 \times {{10}^{ - 19}}\,C,\,\,{m_e} = 9.11 \times {{10}^{ - 31}}\,kg} \right)$$
$$\left( {e = 1.6 \times {{10}^{ - 19}}\,C,\,\,{m_e} = 9.11 \times {{10}^{ - 31}}\,kg} \right)$$
$$2.65 \times {10^6}\,m/s$$
$$7.02 \times {10^{12}}\,m/s$$
$$1.87 \times {10^6}\,m/s$$
$$32 \times {10^{ - 19}}\,m/s$$
Explanation
$$eV = {1 \over 2}m{v^2}$$
$$ \Rightarrow v = \sqrt {{{2ev} \over m}} = \sqrt {{{2 \times 1.6 \times {{10}^{ - 19}} \times 20} \over {9.31 \times {{10}^{ - 31}}}}} $$
$$ = 2.65 \times {10^6}\,m/s$$
$$ \Rightarrow v = \sqrt {{{2ev} \over m}} = \sqrt {{{2 \times 1.6 \times {{10}^{ - 19}} \times 20} \over {9.31 \times {{10}^{ - 31}}}}} $$
$$ = 2.65 \times {10^6}\,m/s$$
Comments (0)
