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Electrostatic Pressure

Electrostatic Pressure :- Electrostatic pressure describes the mechanical force per unit area exerted by an electric field on a charged surface. It arises due to the mutual repulsion of like charges distributed over the surface of a body.

When electric charges accumulate on a conductor, they repel each other due to the electrostatic force. This repulsion creates a kind of “pressure” that tries to push the charges apart, acting outward from the surface. This outward force per unit area is called electrostatic pressure.

Formula for Electrostatic Pressure

Consider an element AB of area dS on the surface of a charged conductor. If σ is the surface charge density on AB, then charge on it is

$\displaystyle dq=\sigma dS$

Now consider two points P and Q just outside and inside of element AB as shown in figure. The whole conductor can be assumed to be made us of two parts AB and ACB. At point P, $\displaystyle {{\overrightarrow{E}}_{AB}}$ represents electric field due to segment AB and $\displaystyle {{\overrightarrow{E}}_{ACB}}$ represents the electric field due to the remaining part of the conductor ACB (if we remove the portion AB then at point P electric field will be produced by the remaining part ACB). Similarly inside the conductor at point Q, electric fields due to the parts AB and ACB are shown.

As $\displaystyle {{\overrightarrow{E}}_{AB}}$ and $\displaystyle {{\overrightarrow{E}}_{ACB}}$ are in the same direction and electric field just outside the conductor is given by $\displaystyle \frac{\sigma }{{{\varepsilon }_{0}}}$ , so net electric field at point P is given by :

$\displaystyle {{E}_{P}}=\left| {{\overrightarrow{E}}_{AB}} \right|+\left| {{\overrightarrow{E}}_{ACB}} \right|$

$\displaystyle {{E}_{P}}={{E}_{AB}}+{{E}_{ACB}}=\frac{\sigma }{{{\varepsilon }_{0}}}$ …..(1)

Also at point Q, we see that $\displaystyle {{\overrightarrow{E}}_{AB}}$ and $\displaystyle {{\overrightarrow{E}}_{ACB}}$ are opposite in direction. Hence net electric field at point Q is given by :

$\displaystyle {{E}_{Q}}={{E}_{AB}}-{{E}_{ACB}}$

But electric field inside a conductor is zero, so

$\displaystyle {{E}_{Q}}={{E}_{AB}}-{{E}_{ACB}}=0$ …..(2)

From equations (1) and (2), we get

$\displaystyle {{E}_{AB}}={{E}_{ACB}}=\frac{\sigma }{2{{\varepsilon }_{0}}}$ …..(3)

Hence the part ACB of the conductor produced an electric field $\displaystyle \frac{\sigma }{2{{\varepsilon }_{0}}}$ . Now this electric field at the location of section AB due to remaining part ACB, exerts an outward force on section AB. This force is given by,

$\displaystyle dF=dq\times {{E}_{ACB}}=\sigma dS\times \frac{\sigma }{2{{\varepsilon }_{0}}}$ …..(4)

Thus electrostatic pressure experienced by the section AB can be given as

$\displaystyle {{P}_{E}}=\frac{dF}{dS}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}$ …..(5)

As net electric field outside the surface of a conductor is $\displaystyle E=\frac{\sigma }{{{\varepsilon }_{0}}}$ , $\displaystyle \Rightarrow \sigma ={{\varepsilon }_{0}}E$ . Using the value of σ in equation (4), we get electrostatic pressure :-

$\displaystyle {{P}_{E}}=\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$ …..(6)

Note :-

If we want to calculate the force experienced by the entire surface of the conductor, then from equation (3), we get

$\displaystyle F=\int\limits_{S}{dF}=\int\limits_{S}{\frac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}dS=}\int\limits_{S}{\frac{{{\varepsilon }_{0}}{{E}^{2}}}{2}dS}$ …..(7)

Electrostatic Pressure

Formula for Electrostatic Pressure

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