Question

TWO Point change s of 20 mc and 80mc are placed 18 cm apart. Find the position where the electric field is zero

Answer 1

ANSWER:

x=6cm

(18-x)=18-6=12cm

GIVEN:

$Q_1 = 20 \times {10}^{ - 3}C \\ Q_2 = 80 \times {10}^{ - 3}C$

r =18cm=0.18m

TO FIND:

x=? , where x is the distance from 20mC charge where electric field is zero

FORMULA:

E =$\frac{1}{4\pi \epsilon_o} \times \frac{Q}{ {r}^{2}}$

EXPLANATION:

$E_1 = E_2 \\ \\ \frac{9 \times {10}^{9} \times 20 \times {10}^{ - 3} }{ {x}^{2} } = \frac{9 \times {10}^{9} \times 80 \times {10}^{ - 3} }{ {(18 - x)}^{2} } \\ \frac{20}{ {x}^{2} } = \frac{80}{ {(18 - x)}^{2} } \\ \frac{1}{ {x}^{2} } = \frac{4}{ {(18 - x)}^{2} } \\ taking \: square \: root \: on \: both \: sides \\ \frac{1}{x} = \frac{2}{18 - x} \\ 18 - x = 2x \\ 3x = 18 \\ x = 6cm$

THE DISATANCE WHERE THE ELECTRIC FIELD IS ZERO IS 6cm FROM THE 20mC CHARGE AND 12cm FROM THE 80mC CHARGE.