The force between two charge when seprate by a distance of 50cm in air is 40 newton what will be the force between them if the distance becomes 25?

158.4N..................m..................,..................................,.........!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Dear, As given is F=kq1q2/r×r= Here we had changed r only So F is inversely proportional to r square. So F1/F2= 25×25/50×50 ( Cancel 50 with 25) F1=40(given) 40/F2=1/4 So F2=160

160 N is answer.....................mmm................................................................mmmmmmm.............................

As the charges are same so they will cancel each other.Therefore,F1/F2=25×25/50×5040/F2=1/4F2=40×4=160N