Question #7b283

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Question #7b283

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Answer 1 · 2017-12-06T06:03:33

I used the formula of $E = k_{e} \frac{q_{1}}{r^{2}}$ $E=k_eq_1/r^2$

Explanation:

This is the Coulumb's Law
$F_{e} = k \frac{q_{1} q_{2}}{r^{2}}$ $F_e=k(q_1q_2)/r^2$
$F_{e} = q E$ $F_e=qE$ ; $\frac{F_{e}}{q} = E$ $F_e/q=E$
This cancels out one of the charges.
Also, $k_{e} = \frac{1}{4 π ε_{o}}$ $k_e=1/(4piepsilon_o)$
$ε_{o} = 8.85 \cdot 10^{- 12} \frac{F}{m}$ $epsilon_o=8.85*10^-12 F/m$ in air
Plugging it in I get
$k_{e} = 8.99 \cdot 10^{9}$ $k_e=8.99*10^9$
Plugging that into our Derived Equation
$E = (8.99 & 10^{9} N m^{2} C^{- 2}) \frac{100 C}{1^{2} m^{2}}$ $E=(8.99&10^9 N m^2 C^-2)(100 C)/(1^2m^2)$
$E = 8.99 \cdot 10^{11} \frac{N}{C}$ $E=8.99*10^11 N/C$

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Question #7b283

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