C) 40o J D) 2.4 x 10\'3J E) -1.5 x 10 8J Problem 8(5.00 points) Using the same f

Question

C) 40o J D) 2.4 x 10'3J E) -1.5 x 10 8J Problem 8(5.00 points) Using the same figure as above, what is the direction of the electric field at point C? A) Straight to the right diagonal down and to the right diagonal up and to the left D) towards the largeconductor E) into the plane of the paper Problem 9(5.00 points) Four ions of equal magnitude charge, +Q are arranged on two opposite corners of a square and- on the other two opposite corners, with sides of length d. What is the net electric field at the center of the square? 4kQ2 A) 2 4kQ2 B) V2d2 The electrie field is zero C) -4kQ2 -4kQ2 d2 Problem 1o(5.o0 points) Assuming a negligible internal resistance in a g.oV battery, that is the change in electric potential energy when-o.5 C of charge is transferred from the negative end to the positive end? A) 45J 9.0J 18 J -4.5J E) -9.0 J

Explanation / Answer

problem 9:

electric field due to positive charge is away from the charge and electric field due to negative charge is directed towards the charge.

in the given configuration, field due to +Q charges on two opposite corners will cancel each other.

similarly, electric field due to -Q charges on two opposite corners will cancel out each other.

so net field will be zero.

hence option C is correct.

problem 10:

electric field is directed from +ve end to -ve end.

as the charge is negative, force is from -ve end to +ve end .

(force=charge*electric field)

so when the charge is moved from -ve end to +ve end, it is moving along the force and work done is positive.

now work done=magnitude of charge*potential difference=0.5*9

=4.5 J

hence option A is correct.

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