Answer:
Ep = 3797.05 N/C in the direction leaving the charge q₂ towards point P
Step-by-step explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/d²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
d: distance from charge q to point P in meters (m)
Data
k= 8.99*10⁹ N*m²/C²
q₁ = +13.6 x 10⁻⁹C
q₂ = +61.0*10⁻⁹C
d₁ =d₂= 33.5 cm = 0.335 m
Look at the attached graphic:
E₁: Electric Field at point P due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge
E₂: Electric Field at point P due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge
E₁ = k*q₁/d₁² = 8.99*10⁹ *13.6 *10⁻⁹/(0.335)² = 1089.45 N/C
E₂ = k*q₂/d₂²=- 8.99*10⁹ *61*10⁻⁹/(0.335)² = - 4886.5 N/C
Magnitude of the electric field at a point midway between q₁ and q₂
Ep= - 4886.5+ 1089.45 = -3797.05 N/C
Ep = 3797.5 N/C in the direction leaving the charge q₂ towards point P