If the radius of a spherical Gaussian surface is doubled, how does the electric
Practice Questions
Q1
If the radius of a spherical Gaussian surface is doubled, how does the electric field due to a point charge at its center change?
It doubles
It halves
It remains the same
It becomes zero
Questions & Step-by-Step Solutions
If the radius of a spherical Gaussian surface is doubled, how does the electric field due to a point charge at its center change?
Step 1: Understand what a point charge is. A point charge is a charged particle that is very small compared to the distance from it to other objects.
Step 2: Know what a spherical Gaussian surface is. It is an imaginary sphere used in physics to help calculate electric fields.
Step 3: Remember that the electric field (E) created by a point charge depends only on the amount of charge (Q) and the distance (r) from the charge, according to the formula E = k * Q / r^2, where k is a constant.
Step 4: Realize that when you double the radius of the Gaussian surface, you are changing the distance (r) from the charge to the surface, but not the charge itself.
Step 5: Understand that the electric field at the surface of the Gaussian sphere is still determined by the charge at the center, not the size of the sphere.
Step 6: Conclude that even if the radius of the Gaussian surface is doubled, the electric field due to the point charge at its center remains the same.
Gauss's Law – The electric field due to a point charge is determined by Gauss's Law, which states that the electric field is proportional to the charge and inversely proportional to the square of the distance from the charge, but is independent of the size of the Gaussian surface.
Electric Field of a Point Charge – The electric field (E) created by a point charge (Q) is given by the formula E = k * Q / r^2, where k is Coulomb's constant and r is the distance from the charge.
