Question

Please See The Attachment​

Answer 1

Answer:

Electrostatic

Explanation:

★ Given ;

» Electric Field ( E) = 90 N / C

» Radius ( r ) =. m

» Magnitude of charge ( q ) = ??

★ We know that ;

Electric Field ( E ) = K q / r²

» q = E × r² / K

» q = 90 × ( 1 )² / 9 × 10⁹

» q = 90 / 9 × 10⁹

» q = 10 × 10-⁹

★ Q ( magnitude of charge ) = 10-⁸ Coulumb

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Answer 2

Given ,

• Electric field (E) = 90 N/c
• Distance (r) = 1 m

We know that , the electric field at point a due to point charge is given by

$\large \boxed{ \sf{E = \frac{k q }{ {(r)}^{2} } }}$

Thus ,

$\sf \mapsto 90 = \frac{9 \times {(10)}^{9} \times q }{ {(1)}^{2} } \\ \\ \sf \mapsto q = \frac{10}{ {(10)}^{9} } \\ \\ \sf \mapsto q = {(10)}^{ - 8} \: \: c$

$\sf \therefore \underline{The \: magnitude \: of \: charge \: is \: {(10)}^{ - 8} \: \: c }$