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?

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 =

Charge =

Distance from the negative charge = 0.05 m

Distance to = 0.032 m

a) Electric field = E =

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

E = =  (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.

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