Question

1, Two charges are placed on the x axis. One of the charges (q1 = +8.09C) is at x1 = +3.00 cm and the other (q2 = -21.5C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

2. Two charges, -22 C and +3.5 C, are fixed in place and separated by 3.0 m. (a) At what spot along a line through the charges is the net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) (b) What would be the force on a charge of +14 C placed at this spot?

Answer 1

1 )

q₁ = + 8.09 C

x₁ = + 3.00 cm = 0.03 m

q₂ = - 21.5 C

x₂ = +9.00 cm = 0.09 m

a )

at x = +3.00 cm

E = - k q₁ / 0.03² + k q₂ / 0.09²

E = - 9 X 10⁹ ( 8.09 / 0.03² + 21.5 / 0.09² )

E = - 1.047 X 10¹⁴ N/C

b )

at x = +6.00 cm

E = k q₁ / 0.03² + k q₂ / 0.09²

E = 9 X 10⁹ ( 8.09 / 0.03² + 21.5 / 0.03² )

E = 2.959 X 10¹⁴ N/C

2 )

a )

given

q₁ = -22 C = | -22 | = 22 C

q₂ = +3.5 C = | + 3.5 | = 3.5 C

d = 3.0

here E₁ = E₂

kq₁ / ( d +x )² = kq₂ / ( x )²

q₁ / ( d +x )² = q₂ / ( x )²

x = q₂² / q₁² + q₂²

x = d / ( ( q₁/ q₂ )^1/2 + 1 )

x = 3 / ( 22/3.5 )^1/2 + 1

x = 0.855 m

at x = 0.855 m the net electric field zero

b )

if the force on a charge of +14 C placed at this spot is also the force is ZERO.