Solve Question : An unknown changes of radius 10 cm isn20 cm away from

1.	Solve Question : An unknown changes of radius 10 cm isn20 cm away from the cnetre. It has E = 1.5 × 10³ find q
Answer» ${\sf{\boxed{Charge\:and\:Electric \:Field}}}$ GIVEN that ✔Distance from OBSERVATION POINT (R) = 20 cm = 20 × $10{}^-2 m$ ✔Electric Field (E) = 1.5 × 10³ N/C We KNOW that ♦ $\frac{kq}{r {}^{2} }$ E = $1.5 \times 10 {}^{3} = \frac{9 \times 10 {}^{9} \times q }{(20 \times 10 {}^{ - 2} ) {}^{2} } \\ \\ \\ = > q = \frac{1.5 \times 10 {}^{3} \times 20 \times 20 \times 10 {}^{ - 4} }{9 \times 10 {}^{9} }$ $= > q = 6.6 \times 10 {}^{ - 9}$ •°• ${\underline{\bold{The\:value\:of\:Charge\:is}}}→$ 6.6 × $10{}^-19$ _______________________________________________

Solve Question : An unknown changes of radius 10 cm isn20 cm away from the cnetre. It has E = 1.5 × 10³ find q

Answer»

${\sf{\boxed{Charge\:and\:Electric \:Field}}}$

GIVEN that

✔Distance from OBSERVATION POINT (R) = 20 cm = 20 × $10{}^-2 m$

✔Electric Field (E) = 1.5 × 10³ N/C

We KNOW that

♦ $\frac{kq}{r {}^{2} }$ E =

$1.5 \times 10 {}^{3} = \frac{9 \times 10 {}^{9} \times q }{(20 \times 10 {}^{ - 2} ) {}^{2} } \\ \\ \\ = > q = \frac{1.5 \times 10 {}^{3} \times 20 \times 20 \times 10 {}^{ - 4} }{9 \times 10 {}^{9} }$