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## At what point the electric field due to a charged spherical shell is maximum?

Thus, the electric field outside a Shell is the same . The charge enclosed inside the spherical shell is 0. Therefore, due to the electric field, the uniformly charged spherical shell is **zero at all points** inside the shell.

## Where is electric field maximum in a sphere?

That when the sphere is at the maximum of E_{} the electrostatic sphere is **a distance a from the center of the sphere** and the charge would be at its maximum. Right, the charge is at its maximum. So while using Gauss’ Law, the q_{enc} is simply Q_{}, and the radius of your Gaussian surface is the radius of the sphere.

## Where the electric field due to a uniform distribution of charge on a spherical shell is zero?

we know that **electric field** of an ideal **spherical shell** with uniformly distributed **charge** is **zero** inside the **shell** and equal to EF of a point **charge** on its center. when we calculate the EF for a point on the surface of **shell**,it is equal to half of EF of same point **charge** EF.

## What is the electric field inside a uniformly charged spherical shell?

Since the charge q is distributed on the surface of the spherical shell, there will be no charge enclosed by the spherical Gaussian surface i.e. = 0. Hence, **there is no electric field inside** a uniformly charged spherical shell.

## Why is the charge inside a spherical shell zero?

The electric field immediately above the surface of a conductor is directed normal to that surface. … Now, the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss’ law, and symmetry, that the **electric field inside the shell is zero**.

## What is the angle between electric field and electric potential?

The angle between the electric field and the equipotential surface is always **90 ^{}**. The equipotential surface is always perpendicular to the electric field.

## Where is the maximum value of electric field intensity due to charged sphere?

Answer Expert Verified

Hence, at **r = 2R/3** electric field is maximum.

## What is the angle between electric dipole moment and electric field strength?

The electric field strength due to an electric dipole at a point in its equatorial position, is parallel to axis of dipole. Also the two are antiparallel, i.e., in opposite directions to each other. Hence, the angle between electric field strength and electric dipole moment is **180∘** .

## What is a test charge in the electric field?

The charge that **is used to measure the electric field strength** is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. … Since electric field is defined as a force per charge, its units would be force units divided by charge units.

## Which of the following is correct the electric field at a point is?

Electric field at a point is continuous if there is no charge at that point. And the field is discontinuous if there is charge at that point. So both options **(b) and (c)** are correct.

## What are the applications of Gauss theorem?

Gauss’s Law can be used **to solve complex electrostatic problems involving unique symmetries like cylindrical, spherical or planar symmetry**. … Gauss’s Law can be used to simplify evaluation of electric field in a simple way.