Answer:
[tex]W=46.08\times 10^{-14}J[/tex]
Explanation:
The work done(W) in bringing 2 protons to a separation 'r' is given as:
[tex]W=\frac{kq^2}{r}[/tex]
where,
k= coulomb's constant = 9 × 10⁹ N
q = charge of protons = 1.6 × 10⁻¹⁹ C
Now, the third charge (or proton) is brought near the other two protons
Thus, work done against both these is
[tex]W_2+W_3=\frac{kq^2}{r}+\frac{kq^2}{r}[/tex]
Now,
The total work done (W) = [tex]W_1 +W_2+W_3=3\frac{kq^2}{r}[/tex]
or
[tex]W=3\times \frac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.5\times 10{-15}}[/tex]
or
[tex]W=46.08\times 10^{-14}J[/tex]