Draw the FBD. Let the length of the thread be L and the separation between the masses be x. Also assume the angle made by the string to the vertical to be . Then

equations of motion

T-mgcos-(kq²/x²)sin=ma_n

-mgsin+(kq²/x²)cos = ma_t

where a_n and a_t are the normal and tangential acceleration components. Now at the moment of release v=0, so a_n = (v²/R) = 0

Also note that I have taken positive tangential direction away from the other charge.

So net force alon the string, ma_n = 0

Net force perpendicular to the string, ma_t = -mgsin+(kq²/x²)cos

Substitute the values, the trigonometric ratios can be found from the diagram

You can find the tension from the first equaion.

At the initial moment a = a_t = -gsin+(kq²/mx²)cos

from the FBD write the eqns  Fq - tsin theta  = ma    ----------------1

                                                    --------------2

                                                   T cos theta = .98 --------------------3

Fq = k  ( 2 x 10 ^{- 7})² /(.05)²    =   .144 N

from the diagram cos theta = .499 / .500       ( use pythagoras theoram to find the  vertical dis)

use this in 3 to find T that comes to . 9816 N

use the value of T in  eqn 1 u get the  value of a = .95

electric force = f= kQ1Q2/d SQUARE.............= TO 0.144N

2ND PART   IS     F =MG =TSIN THETA

C PART IS               T SIN THETA =F

                                     TCOS THETA =MG

                                     T SQUARE =F SQUARE + MG SQUARE

                                     T=0.986N ANS