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Electric Potential

Electric Potential :- Electric potential, often denoted as V, is the amount of electric potential energy per unit charge at a point in an electric field. In simpler terms, it tells us how much work is needed to move a unit positive charge from a reference point (usually infinity) to a specific point in the electric field.

Definition of Electric Potential

“The electric potential at a point in an electric field is defined as the amount of work done per unit positive test charge by an external force in bringing it from infinity to that point against the field without acceleration.”

Formula for Electric Potential

If an amount of work W_∞P is done to move a positive test charge q_o from infinity to a point P against the electric field, then formula for electric potential at point P is :

$\displaystyle V=\frac{{{\left( {{W}_{\infty P}} \right)}_{ext.}}}{{{q}_{o}}}$

The SI unit of electric potential is the volt (V). One volt equals one joule per coulomb :

$\displaystyle 1V=\frac{1J}{1C}$

Definition of 1 Volt :- “If one joule of work is done in moving one coulomb of positive charge without acceleration from infinity to a point in the electric field against the field by a external force, then potential at that point is called one volt.”

Dimensions :-

$\displaystyle \left[ V \right]=\left[ \frac{W}{q} \right]=\frac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ AT \right]}$

$\displaystyle \therefore \left[ V \right]=\left[ M{{L}^{2}}{{T}^{-3}}{{A}^{-1}} \right]$

Note :-

(i) Potential is a scalar quantity, its value may be positive, negative or zero.

(ii) Electric potential at a point is also equal to the negative of the work done by the electric field in taking the point charge from reference point (i.e. infinity) to that point.

$\displaystyle V=\frac{-{{\left( {{W}_{\infty P}} \right)}_{Electric\text{ }Field}}}{{{q}_{o}}}$

(iii) Electric potential due to a positive charge is always positive and due to negative charge it is always negative except at infinity. (taking V_∞ = 0).

(iv) If a charge particle is released from rest in an electric field, then work done by electric field on the charge particle is equal to gain in it’s kinetic energy.

$\displaystyle {{\left( {{W}_{P\infty }} \right)}_{Electric\text{ }Field}}=-{{\left( {{W}_{\infty P}} \right)}_{Electric\text{ }Field}}=-\left[ -{{\left( {{W}_{\infty P}} \right)}_{ext.}} \right]={{\left( {{W}_{\infty P}} \right)}_{ext.}}$

$\displaystyle \Rightarrow {{\left( {{W}_{P\infty }} \right)}_{Electric\text{ }Field}}=q{{V}_{P}}$

Also from work energy theorem,

$\displaystyle {{\left( {{W}_{P\infty }} \right)}_{Electric\text{ }Field}}=\Delta K={{K}_{\infty }}-{{K}_{P}}={{K}_{\infty }}-0={{K}_{\infty }}$

Hence,

$\displaystyle {{\left( {{W}_{P\infty }} \right)}_{Electric\text{ }Field}}={{K}_{\infty }}=q{{V}_{P}}$

(v) Potential decreases in the direction of electric field.

Example 1.

A charge of 2μC is taken from infinity to a point in an electric field, without changing its velocity. If work done against electrostatic forces is –40μJ, then find the potential at that point.

Solution :

$\displaystyle V=\frac{{{\left( {{W}_{\infty P}} \right)}_{ext.}}}{{{q}_{o}}}=\frac{-40\mu }{2\mu }=-20Volt$

Example 2.

When charge 10 μC is shifted from infinity to a point in an electric field, it is found that work done by electrostatic forces is –10 μJ. If the charge is doubled and taken again from infinity to the same point without accelerating it, then find the amount of work done by electric field and against electric field.

Solution :

$\displaystyle V=\frac{-{{\left( {{W}_{\infty P}} \right)}_{Electric\text{ }Field}}}{{{q}_{o}}}=\frac{-\left( -10\mu \right)}{10\mu }=1Volt$

So, if now the charge is doubled and taken from infinity to the same point without acceleration then,

$\displaystyle V=\frac{{{\left( {{W}_{\infty P}} \right)}_{ext.}}}{{{q}_{o}}}\text{ }\Rightarrow \text{ 1}=\frac{{{\left( {{W}_{\infty P}} \right)}_{ext.}}}{20\mu }$

So work done by external force,

$\displaystyle {{\left( {{W}_{\infty P}} \right)}_{ext.}}=20\mu J$

Similarly work done by electric field,

$\displaystyle {{\left( {{W}_{\infty P}} \right)}_{Electric\text{ }Field}}=-{{\left( {{W}_{\infty P}} \right)}_{ext.}}=-20\mu J$

Example 3.

A charge 3μC is released from rest from a point P where electric potential is 20 V then find its kinetic energy when it reaches infinity.

Solution :

$\displaystyle {{\left( {{W}_{P\infty }} \right)}_{Electric\text{ }Field}}={{K}_{\infty }}=q{{V}_{P}}=(3\mu )\left( 20 \right)=60\mu J$

Electric Potential

Definition of Electric Potential

Formula for Electric Potential

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