Four point charges q_A = 2 μC, q_B = – 5 μC, q_c = 2 μC, and q_D = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

Hence, two field lines never cross each other.

                          
and                        
(a) 
         
Let us assume that a unit positive test charge is placed at 0. q_A will repel this test charge while q_B will attract. Hence,  both are directed towards 
                           

                               
                               
                                     
(b) As a negative test charge of q₀ = – 1.5 x 10^–6 C is placed at 0. q_A will attract it while q_B will repel. Therefore, the net force




   

Given,

 $$\begin{gathered} \mathrm{Dipole} \mathrm{moment}, \mathrm{p} = 4 \times {10}^{-9} \mathrm{Cm}; \\ \mathrm{\theta } = 30° \\ \mathrm{Eelectric} \mathrm{field}, \mathrm{E} = 5 \times {10}^4 {\mathrm{NC}}^{-1} \end{gathered}$$

Torque is given by  

      $$\begin{gathered} \tau =\boxed{\mathrm{p}\times \overrightarrow{\mathrm{E}}} = \mathrm{p}. \mathrm{E} \sin \mathrm{\theta } \\ = 4 \times {10}^{-9} \times 5 \times {10}^4 \times \sin 30° \end{gathered}$$
    
                $\tau =4\times {10}^{-9}\times 5\times {10}^4\times \frac{1}{2}$
                   $\tau ={10}^{-4} \mathrm{Nm}$