Question

what ia the net direction of net coulomb force on Q3 from + x axis? (answer in degree).

Q2 is 50cm to the right from Q1 and Q3 is 40cm below Q1.

Q1 = 35 x 10-6 C

Q2= 65 x 10-6 C

Q3 = 80 x 10-6 C

Answer 1

As per the given question,

$Q_1=35 \times10^{-6 }C$

$Q_2=65 \times 10^{-6} C$

$Q_3=80 \times10^{-6} C$

$F_{21}=\frac{Q_1Q_2}{4\pi \epsilon_o r_1^2}$

Now, substituting the values in the above,

$\Rightarrow F_{12}=\frac{35\times 10^{-6}\times {65\times 10^{-6}\times 9\times 10^{-9}}}{4\pi \epsilon_o {0.5}^2}\hat{(i)}$

Similarly,

$F_{23}=\frac{Q_2Q_3}{4\pi \epsilon_o r_2^2}$

$\Rightarrow F_{23}=\frac{80\times 10^{-6}\times {65\times 10^{-6}\times 9\times 10^{-9}}}{4\pi \epsilon_o {0.4}^2}\hat{(-j)}$

$\tan\theta=\frac{F_{(23)}}{F_{(21)}}$

$\Rightarrow \tan\theta=\frac{80\times 0.4\times 0.4}{35\times 0.5\times 0.5}$

$\Rightarrow \tan\theta =\frac{16\times 16}{35\times 5}$

$\theta = 55.64^\circ$