Question

Solve Question : An unknown changes of radius 10 cm isn20 cm away from the cnetre. It has E = 1.5 × 10³ find q
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Answer 1

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

Given that

✔Distance from observation point (r) = 20 cm = 20 × $10{}^-2 m$

✔Electric Field (E) = 1.5 × 10³ N/C

We know that

♦E = $\frac{kq}{r {}^{2} }$

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

$= > q = 6.6 \times 10 {}^{ - 9}$

•°• ${\underline{\bold{The\:value\:of\:Charge\:is}}}→$ 6.6 × $10{}^-19$

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Answer 2

Explanation:

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