Home / AP Physics C E&M – 8.3 Electrostatics: Electric Field and Electric Potential- Exam Style questions- MCQs
Two small spheres are arranged along a line and carry charges of \(+4Q\) and \(-3Q\), as shown in the figure above. The vertical lines are equally spaced. At which of the labeled points does the electric field point toward the right with the smallest magnitude?
Correct Answer: \( \boxed{\mathrm{C}} \)
The electric field due to a positive charge points away from the charge, while the field due to a negative charge points toward the charge.
Between the two charges, both electric fields point toward the right, so their magnitudes add together. To the left of both charges or to the right of both charges, the fields oppose each other.
The electric field magnitude from a point charge is
\(E = k\dfrac{|q|}{r^{2}}.\)
Comparing the labeled points, point \(C\) lies between the charges where the net electric field is directed to the right and has the smallest magnitude among the locations where the field points right.
Therefore, the correct answer is (C).
Three particles each with charge \(+Q\) are placed on three corners of a square, as shown above. The sides of the square have length \(s\). Point C is at the center of the square. What is the direction of the electric field at point C?
Correct Answer: \( \boxed{\mathrm{E}} \)
The electric field due to a positive charge points away from the charge.
Since point C is at the center of the square, each of the three charges is the same distance from C. Therefore, each charge produces an electric field of equal magnitude:
\(E=k\dfrac{Q}{r^{2}}.\)
The fields due to the charges at the upper-left and lower-left corners have equal and opposite vertical components, which cancel. Their horizontal components both point to the right.
The field due to the charge at the upper-right corner points diagonally toward the lower-left. Adding this vector to the resultant of the two left charges leaves a net field directed diagonally downward and to the right.
Therefore, the electric field at point C points in the direction shown in (E).
What is the radial component of the electric field associated with the potential
\( V=ar^{-2} \)
where \(a\) is a constant?
Correct Answer: \( \boxed{\mathrm{E}} \)
The radial component of the electric field is related to the electric potential by
\( E_{r}=-\dfrac{dV}{dr} \)
Given
\( V=ar^{-2} \)
Differentiate with respect to \(r\):
\( \dfrac{dV}{dr}=a(-2r^{-3})=-2ar^{-3} \)
Therefore,
\( E_{r}=-\left(-2ar^{-3}\right)=2ar^{-3} \)
Thus, the radial component of the electric field is
\( \boxed{E_{r}=2ar^{-3}} \)
Therefore, the correct answer is (E).