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 ?
$\left(e=1.6\times10^{-19}C, m_{e}=9.11\times10^{-31}\,kg\right)$
Since $V_2 > V_1$, so electric field will point from plate 2 to plate 1.
The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.
Use work-energy theorem to find speed of electron when it strikes the plate 2.
$\frac{m_{e}V^{2}}{2}-0=e\left(V_{2}-V_{1}\right)$
where $v$ is the required speed.
$\therefore \frac{9.11\times10^{-31}}{2}v^{2}=1.6\times10^{-19}\times20$$\Rightarrow v=\sqrt{\frac{1.6\times10^{-19}\times40}{9.11\times10^{-31}}}=2.65\times10^{6}\,m/s$
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Top Questions on electrostatic potential and capacitance
The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.
Some major things that we should know about electric potential:
They are denoted by V and are a scalar quantity.
It is measured in volts.
Capacitance
The ability of a capacitor of holding the energy in form of an electric charge is defined as capacitance. Similarly, we can also say that capacitance is the storing ability of capacitors, and the unit in which they are measured is “farads”.
The capacitor is in Series and in Parallel as defined below;
In Series
Both the Capacitors C1 and C2 can easily get connected in series. When the capacitors are connected in series then the total capacitance that is Ctotal is less than any one of the capacitor’s capacitance.
In Parallel
Both Capacitor C1 and C2 are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is Ctotal is any one of the capacitor’s capacitance.
