An electric dipole shown in the figure. Work done to move a charge particle of 1 C from point Q to P is x 10⁻⁷ ~J , then the value of x is ..........

Step-by-step solution

From figure, electric dipole moment, aligned p & =2 q l=q(2 l) & =2 10⁻⁶ (1 10⁻⁹ ) p & =2 10⁻¹⁵ ~cm aligned Electric potential at point p due to dipole aligned V_p & = 1 4 ₀ p r^2 V_p & =9 10^9 2 10⁻¹⁵ 60^ (1 10⁻² )^2 & = 18 10⁻⁶ 1 2 10⁻⁴ =9 10⁻²=0.09 ~V aligned Similarly, electric potential at point Q due to electric dipole aligned V_Q & = 9 10^9 (2 10⁻¹⁵ ) 120^ (1 10⁻² )^2 & =-0.09 ~V aligned Potential difference between point Q and P . V_ Q P =V_P-V_Q=0.09-(-0.09)=0.18 ~V Work done to move the charge of 1 C from