If the radius of a charged sphere is halved while keeping the charge constant, w

If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?

If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?

Step 1: Understand that the electric field (E) at the surface of a charged sphere is calculated using the formula E = Q/(4πε₀R²), where Q is the charge, ε₀ is a constant, and R is the radius of the sphere.
Step 2: Note that in this problem, the charge (Q) remains constant while the radius (R) is halved.
Step 3: When the radius (R) is halved, we can express the new radius as R/2.
Step 4: Substitute R/2 into the formula for electric field: E' = Q/(4πε₀(R/2)²).
Step 5: Simplify the equation: E' = Q/(4πε₀(R²/4)) = Q/(4πε₀R²) * 4 = 4 * (Q/(4πε₀R²)).
Step 6: This shows that the new electric field (E') is 4 times the original electric field (E).
Step 7: Conclude that when the radius of the sphere is halved, the electric field at the surface increases by a factor of 4.

Electric Field of a Charged Sphere – The electric field at the surface of a charged sphere is inversely proportional to the square of its radius, given by the formula E = Q/(4πε₀R²).