due to −Q points toward −Q. Since these vectors point in the same direction,
the net electric field at P is (E to the right) + (E to the right) = (2E to the
right).
- D The acceleration of the small sphere is
As r increases (that is, as the small sphere is pushed away), a
decreases. However, since a is always positive, the small sphere’s
speed, v, is always increasing.
- B Since FE (on q) = qE, it must be true that FE (on −2q) = −2qE = −2FE.
- D All excess electric charge on a conductor resides on the outer surface.
- D By definition, electric field vectors point away from a positive source
charge and toward a negative source charge. Furthermore, since an electron
(which is negatively charged) would be repelled from a negative source
charge, the resulting electric force on an electron would point away from a
negative source charge, in the opposite direction from the electric field
vector.
- A The individual electric field vectors at P due to the two source charges have
the same magnitude and point in opposite directions. Therefore, the net
electric field at point P will be zero.
- A The individual electric field vectors at the center of the square due to the
two negative source charges cancel each other out. So we simply need to
make sure that the individual electric field vector at the center of the square
due to the bottom-right source charge cancels the individual electric field
vector at the center of the square due to the upper-left source charge. Since
the upper-left source charge is +Q, the bottom-right source charge must be
+Q also.
