Question

Four equal charges are located at the corners of a square of side d. If the magnitude of each charge is doubled, then what happens to the resultant force on each charge? It is doubled. It is quadrupled. It increases by a factor of 8. It remains the same. None Of the other choices is correct. A particle travelling along the +x-axis enters an electric field directed vertically upward along the +y-axis. If the particle experiences a force downward because of this field, then what could be said about this particle? It is negatively charged. It is positively charged. It is electrically neutral. It is an electric dipole. None of the other choices is correct. Gaussian surfaces A and B enclose the same positive charge +Q. The area of Gaussian surface A is three times larger than that of Gaussian surface B. The flux of electric field through Gaussian surface A is nine times larger than the flux of electric field through Gaussian surface B. three times larger than the flux of electric field through Gaussian surface B. equal to the flux of electric field through Gaussian surface B. three times smaller than the flux of electric field through Gaussian surface B. unrelated to the flux of electric field through Gaussian surface B. A conductor is placed in an electric field under electrostatic conditions. Which of the following statements is correct for this situation? The electric field is zero inside the conductor. All valence electrons go to the surface of the conductor. The electric field on the surface of the conductor is perpendicular to the surface. All of the other answers apply. None of the other answers apply. An equipotential surface must be parallel to the electric field at any point. perpendicular to the electric field at any point. equal to the electric field at any point. equal to the electric field times distance at any point. unrelated to the electric field.

Answer 1

1. F = a k q1 q2 / d^2

q1 and q2 are doubled.

so Force will quadrupled.

2. vector F = q vectorE

F and E are in opposite direction.

so q is negatively charges.

3.
Using Gauss Law

flux = Qinside / e0 = Q/e0
equal.

4.
field inside the conductor is always zero.

charges resides on surface.

field perpendiculr to the surface.

5.equipotential surface = potential is same everywhere on surface

and perpendicular to the field at any point.