Charge Contained in Symmetrical Objects: Calculation, Examples, Application

Have you ever thought of how the electric field intensity will vary because of the presence of a point charge or charged spherical shell? Do you know that the field inside a charged conducting sphere is zero? All these can be deduced by using the symmetry through Gauss's Law.

Table of content:

• Calculation of contained Charge inside a regular shaped object
• Calculation of electric field due to a point charge
• Practice Problems
• FAQs

Calculation of Charge contained inside a regular shaped object:

When we know the electric field around a regular shaped object, the calculation of the contained charge becomes easy. It can be done by the application of Gauss Law.

According to the Gauss law,

We know,

So,

Here it is to be noted, for ease of calculation the electric field should always be perpendicular or parallel to the area. As it is a dot product either zero or just a simple multiplication will ease off the calculation part.

Calculation of electric field due to a point charge

Gauss law can be useful in determining the electric field due to a point charge at a distance from the point charge. If we consider a sphere as shown, we will find that at every point on that sphere the electric field is uniform. Thus, we can select this sphere as the Gaussian surface. The Electric field is perpendicular to the surface of the sphere and radially outward for a positive point charge.

Here we are to calculate the electric field intensity at a distance from the positive point charge . So we imagine a spherical gaussian surface of radius , centered around the charge. 

The electric field will be perpendicular to the sphere. 

According to the Gauss law,

would have been inward in case the charge contained in it was negative.

Practice Problems:

Q 1. A cube of side , is placed with one of its corners at the origin, as shown. The electric field around the cube is calculated as . Find the charge inside the cube?

A. Here, 

It means, 

Q 2. A point charge of is at the origin. Calculate the electrical field at using Gauss’s Law.

A. Let’s consider an imaginary sphere of radius . 

According to the Gauss law,

Q 3. A cube of side , placed with one of its corners at the origin, as shown. The electric field around the cube is calculated as . Find the charge inside the cube?

A. Here, 

+

It means, 

Q 4. A point charge of is at the origin. Calculate the electrical field at using Gauss’s Law.

A. Let’s consider an imaginary sphere of radius . 

According to the Gauss law,

FAQs:

Q 1. What are the conditions on a Gaussian surface for easy calculation?
A. A gaussian surface can be imagined where the electric field is either parallel or perpendicular to the corresponding surface vector. This makes the cosines in all the dot products equal to one or simply zero. The electric field that passes through the parts of the Gaussian surface where the flux is non-zero should have a constant magnitude.

Q 2. How to determine a Gaussian surface?
A. We need to be clear on what we want. We need to identify whether a surface for a body is a suitable Gaussian surface or not. Therefore, we should check the angle at all points on the surface. Also, the electric field that passes through the parts of the Gaussian surface where the flux is non-zero should have a constant magnitude.

Q 3. A cube of side , placed with one of its corners at the origin, as shown. The electric field around the cube is calculated as . Find the charge inside the cube?

A. Here, 

It means, 

Q 4. A point charge of is at . Calculate the electrical field at using Gauss’s Law.

A. Let’s consider an imaginary sphere of radius . 

According to the Gauss law,