Question

The two charges 3.2×10-19c and -3.2×10-19c are placed 2.4 a apart from an electric dipole it is placed in a uniform electric field of intensity

Answer 1

The value of electric field intensity is 4.9 x 10^-10 N / C

Explanation:

Given data:

q1 = 3.2×10-19 C

q2 = -3.2×10-19 C

Distance between charged "d" 2.4 m

As we know that

F = Kq1q2 / d^2

E = F / Q

E =  Q1 / 4πεod^2

E = 3.2×10-19 / 4 x 3.14 x 8.854 × 10^-12 (2.4)^2

E = 3.2 x 10^-19+12 / 4 x 3.14 x 8.854 x 5.76

E = 3.2 x 10^-7 / 640.55

E = 0.0049 x 10^-7

E = 4.9 x 10^-3-7 = 4.9 x 10^-10 N / C

Thus the value of electric field intensity is 4.9 x 10^-10 N / C

Also learn more

Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2*10-6 C/ m ?

https://brainly.in/question/3734396

Answer 2

Answer:

U=−3×10−23J

Explanation:

Potential energy of electric dipole

U=−pEcosθ=−(q×2l)Ecosθ

U=−(3.2×10−19×2.4×10−10)4×105cosθ

U=−3×10−23J (approx.)