Shakti ka vimiy sutra hai​

E=mc^

i hope answer is correct bro mark me has brainleast

known :

qa = 3x10⁻⁸ c

qb = -6x10⁻⁸ c

qc = 2x10⁻⁸ c

rab = 3 cm = 3x10⁻² m

rbc = 6 cm = 6x10⁻² m

asked: f at point b?

answered:

coulomb force of charge a and b

fab = k qaqb / rab²

= 9x10⁹ x 3x10⁻⁸ x 6x10⁻⁸ / (3x10⁻²) ²

= 9x10⁹ x 18x10⁻¹⁶ / 9x10⁻⁴

= 18x10⁻³ n

coulomb force charge c and b

fbc = k qbqc / rbc²

= 9x10⁹ x 6x10⁻⁸ x 2x10⁻⁸ / (6x10⁻²) ²

= 108x10⁻⁷ / 36x10⁻⁴

= 3x10⁻³ n

so the coulomb force at point b = fab - fbc

= 18x10⁻³ - 3x10⁻³

= 15x10⁻³ n

= 1.5x10⁻² n

answer:

what is the