If the radius of a spherical Gaussian surface is doubled, how does the electric field change if the enclosed charge remains constant?
Practice Questions
1 question
Q1
If the radius of a spherical Gaussian surface is doubled, how does the electric field change if the enclosed charge remains constant?
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, how does the electric field change if the enclosed charge remains constant?
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) due to a point charge is given by the formula E = k * Q / r^2, where k is a constant, Q is the charge, and r is the distance from the charge.
Step 3: Note that if the radius of the Gaussian surface is doubled, the new radius becomes 2r.
Step 4: Substitute the new radius into the formula: E_new = k * Q / (2r)^2.
Step 5: Simplify the equation: E_new = k * Q / (4r^2).
Step 6: Compare the new electric field (E_new) with the original electric field (E): E_new = (1/4) * E, which means the electric field is now one-fourth of the original value.
Step 7: Since the electric field decreases with the square of the distance, if the radius is doubled, the electric field is reduced to half of its original value.
