Question

A charge of 5C is placed at the centre of a spherical Gaussian surface of radius 5cm. The electric flux through the surface is ​​$\dfrac{1}{{{\varepsilon _0}}}$ times of-

Answer 1

In this question, we will use the Gauss’s law equation, which gives relation between electric field, area and total charge. By substituting the given values, we will get the required result. Also, we will discuss the basics of the electric field, for better understanding.

Formula used:
$\int {E.da} = \dfrac{{{Q_{total}}}}{{{\varepsilon _0}}}$

Complete answer:

As we know, Gauss's law is also known as Gauss's flux theorem. Gauss’s law is a law which gives relation between the distributions of electric charge with the resulting electric field.

As we know, that according to Gauss law:

$\int {E.da} = \dfrac{{{Q_{total}}}}{{{\varepsilon _0}}}$

$\Rightarrow \phi = \dfrac{{{Q_{total}}}}{{{\varepsilon _0}}}$

Now, when we substitute the given value of the total charge in the above derived equation, we get:

$\therefore \phi = \dfrac{5}{{{\varepsilon _0}}}$

Therefore, we get the required expression for electric flux, associated with the given charge.

Additional information: As we know, the electric field is defined as the electric force per unit charge:

$E = \dfrac{F}{q}$

Here, F is the force experienced by the given object and q is the charge on the given object. So, here when the given electric field at a particular point is known and force is also known then we can get the charge or if the charge is known to us we can easily get the force experienced by the point charge.

Also, the electric field is radially outward from a positive charge and it is radially inward to a given negative charge. The S.I unit of electric field is known as Newton per coulomb.

Also, the direction of electric field E is taken to be the direction of force it would exert on the positive test charge.

Below is the given diagram of charge flowing in a conductor with electric field direction.

Also, we know that the electric field is defined as the negative of gradient potential (V), and given by:

$E = - gradV$

As we know that the surface of the charged conductor has a constant potential i.e., V. Therefore, there will be no field along the surface of the charged conductor.

Also, potential represented by V, is defined as the amount of work needed to move the given one unit of electric charge from a reference point to a specific point in the presence of an electric field. The S.I unit of potential is given by Volts or joules per coulomb.

Note:
Electric field is a vector quantity. It has both magnitude and direction. Electric field is never negative; rather the negative sign of the electric field shows the opposite direction of the field. Electric field inside a conductor is always zero, this is because charges are not present inside a given conductor, and charges are present on the surface of the conductor.