Question

a point charge A of +25 esu is placed between two charges B and C on the same line with them.The amount of charges in B and C are +5esu and -30esu respectively. The distance between the charges A and B is 2.5 cm and between A and C is 5 cm . what force will be act on the charge ?​

Answer 1

force on charge A is due to A--B and A---C

B-------->--A------>-----C

F(AB)=k25×5/(2.5 ×10^-2)^2

F(AC)=k25×30/(5×10^-2)^2

Now Add the forces since they are in the same direction

So Net Magnitude of force on A will be :-

|F(AB)|+|F(AC)| =

k25(5/(2.5 ×10^-2)^2+30/(5×10^-2)^2)

Solve To get the answer .

[Here I have taken only the magnitude and direction has been take as same so plus.

But if otherwises if magnitude isn't considerd then F(AC) Will be Negative.]