Question

Two pont charges, q1 " 1.6 × 10-7 c and q2 "-6.8 × 10-s c, are held 37.0 ㎝ apart. (Assume q2 is on the right.) (a) What is the electric field (in N/C) at a point 5.0 cm from the neg ative charge and along the line between the two charges? magnitude directionto the right D N/C (b) What is the force (in N) on an electron placed at that point?

Answer 1

a)2.59 ×10⁵ N/C

Direction is to the right.

b) 4.14 × 10⁻¹⁴ N , direction is to the Left.

Explanation:

Given:

Charge
`q_(1)` =
`1.6* 10^(-7)C`

Charge
`q_(2)` =
`-6.8* 10^(-8)C`

Distance from the negative charge = 0.05 m

Distance to
`q_(1)` = 0.032 m

a) Electric field = E =
`k[(q_(1))/(r_(1)^(2)) + (q_(2))/(r_(2)^(2))]`

k = Coulomb's constant = 9 × 10⁹ N m²/C²

E =
`(9* 10^(9))[(1.6 * 10^(-7))/((0.032)^(2)) + (6.8 * 10^(-8))/((0.05)^(2))]` = (1.40625 × 10⁴ + 24.48 × 10⁴ ]

= 2.59 ×10⁵ N/C

Direction is to the right.

b) Force on the electron = Charge on the electron × Electric field E.

⇒ F = ( 1.6 × 10⁻19)(2.59 ×10⁵) = 4.14 × 10⁻¹⁴ N , direction is to the Left.