Question

Find the magnitude of the electric field at a point midway between two charges +13.6 x 10-9 C and +61.0 x 109 C separated by a distance of 67.0 cm. The value of Coulomb's constant is 8.99 x 109 N m2/C2

Answer 1

Answer:

Ep =  3797.05  N/C  in the direction leaving the charge q₂ towards point P

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