Two charges are placed on the x axis. one of the charges (q1 = +7.7 µc) is at x1 = +3.1 cm and the other (q2 = -19 µc) is at x2 = +8.9 cm. (a) find the net electric field (magnitude and direction) at x = 0 cm. (use the sign of your answer to indicate the direction along the x-axis.) n/c (b) find the net electric field (magnitude and direction) at x = +5.5 cm. (use the sign of your answer to indicate the direction along the x-axis.) n/c

a)

The direction will be negative direction.

b)

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

Now by putting the va;ues

a)

The net electric field

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

The net electric field

The direction will be positive direction.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

as technology advances, new experiments often expose problems in accepted theories. c

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-lavira: )