Question

1. A point charge of 4 µC is located at x = -3.0 cm, and a second point charge of -5 µC is located at x = +4.0 cm. Where should a third charge of +6.0 µC be placed so that the electric field at x = 0 is zero?

2. An electron is released from rest in a weak electric field given by -1.7 x 10^-10 N/C . After the electron has traveled a vertical distance of 1.3 µm, what is its speed? (Do not neglect the gravitational force on the electron.)

3. A point charge of -2 µC is located at the origin. A second point charge of 6 µC is at x = 1 m, y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium.

4.Three charges, each of magnitude 4 nC, are at separate corners of a square of edge length 1 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q = +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the +y axis is directed upward.)

Answer 1

Given q = 4 x10 ^-6 C

q ' = -5 x10 ^-6 C

Third charge Q = 6 x10 ^-6 C

Let the third charge is placed at a distance r from x=0 .

Electric field at x= 0 due to charge q is E = Kq/(3cm) ²

= (8.99x10 ⁹)(4x10 ^-6) /(0.03) ² Since 3 cm = 0.03 m

= 3.9955 x10 ⁷ N/C \

It is along positive X axis

Electric field at x= 0 due to charge q ' is E ' = Kq ' /(4cm) ²

= (8.99x10 ⁹)(5x10 ^-6) /(0.04) ²   Since 4 cm = 0.04 m

= 2.8093 x10 ⁷ N/C

It is along positive X axis.

Electric field at x= 0 due to charge Q is E " = KQ/r ²

= (8.99x10 ⁹)(6x10 ^-6) /r ²   

= 53940 / r ²

We know E " = E + E '

( 53940 / r ²) = (3.9955 x10 ⁷ N/C)+(2.8093 x10 ⁷ N/C)

= 6.8048x10 ⁷

r ² = 53940 /(6.8048x10 ⁷)

= 7.926x10 ^-4

r = 2.815x10 ^-2 m

= 2.815 cm

But to get zero electric field r =+2.815 cm