If the radius of a spherical Gaussian surface is doubled while keeping the charge inside constant, how does the electric field change?
Practice Questions
1 question
Q1
If the radius of a spherical Gaussian surface is doubled while keeping the charge inside constant, how does the electric field change?
It doubles
It halves
It remains the same
It quadruples
The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
Questions & Step-by-step Solutions
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Q
Q: If the radius of a spherical Gaussian surface is doubled while keeping the charge inside constant, how does the electric field change?
Solution: The electric field E due to a point charge decreases with the square of the distance from the charge, so if the radius is doubled, the electric field halves.
Steps: 7
Step 1: Understand that we have a spherical Gaussian surface around a point charge.
Step 2: Recall that the electric field (E) created by a point charge decreases with distance from the charge.
Step 3: Remember the formula for the electric field due to a point charge: E = k * Q / r^2, where k is a constant, Q is the charge, and r is the distance from the charge.
Step 4: Note that if we double the radius (r) of the spherical surface, we are increasing the distance from the charge.
Step 5: Since the radius is doubled, we replace r with 2r in the formula: E = k * Q / (2r)^2.
Step 6: Simplify the equation: E = k * Q / (4r^2). This shows that the electric field is now one-fourth of what it was at the original radius.
Step 7: Since the electric field is inversely proportional to the square of the distance, doubling the radius results in the electric field being halved.
