Question:

An electric dipole is along a uniform electric field: If it is deflected by $60^\circ$, work done by an agent is $2 \times 10^{-19}$ Then the work done by an agent if it is deflected by $30^0$ further is

Updated On: Jun 23, 2023

$2.5 \times 10^{-19}$ J
$2 \times 10^{-19}$ J
$4 \times 10^{-19}$ J
$2\times 10^{-16}$ J

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The Correct Option is B

Solution and Explanation

Working on electric dipole when deflected by an angle of $60^{\circ}$ is given by, $W _{1}= U =- PE \cos 60^{\circ}=-2 \times 10^{-19} J .$ Now, work done in deflecting he dipole by another $30^{\circ}$ is given by, $W _{1}=- PE \cos 90^{\circ}=0$ Therefore work done by the agency which deflected dipole by $30^{\circ}$ mpore is, $W = W _{2}- W _{1}=0-\left(-2 \times 10^{-19}\right)=2 \times 10^{-19} J .$

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Electrostatic Potential

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 C_totalis 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 C_total is any one of the capacitor’s capacitance.