If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?
Practice Questions
1 question
Q1
If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?
It remains the same
It doubles
It halves
It quadruples
The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.
Questions & Step-by-step Solutions
1 item
Q
Q: If the radius of a charged sphere is halved while keeping the charge constant, what happens to the electric field at the surface?
Solution: The electric field at the surface of a sphere is given by E = Q/(4πε₀R²). If R is halved, E increases by a factor of 4.
Steps: 7
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.
