Question

question 22 of 25
what is the strength of the electric field of a point - charge of magnitude - 4.8×10^(-19) c at a distance of 4.0×10^(-3) m?
(e = \frac{kq}{r^{2}}, k = 9.00×10^{9} n\cdot m^{2}/c^{2})

a. 3.6×10^(-4) n/c
b. 2.7×10^(-4) n/c
c. - 2.7×10^(-4) n/c
d. - 3.6×10^(-4) n/c

Explanation:

Step1: Identify given values

$k = 9.00\times10^{9}\ N\cdot m^{2}/C^{2}$, $q=- 4.8\times10^{-19}\ C$, $r = 4.0\times10^{-3}\ m$

Step2: Substitute values into formula

$E=\frac{k|q|}{r^{2}}=\frac{9.00\times 10^{9}\times| - 4.8\times10^{-19}|}{(4.0\times10^{-3})^{2}}$

Step3: Calculate the numerator

$9.00\times 10^{9}\times4.8\times10^{-19}=43.2\times10^{-10}=4.32\times 10^{-9}$

Step4: Calculate the denominator

$(4.0\times10^{-3})^{2}=16\times10^{-6}$

Step5: Calculate the electric - field strength

$E=\frac{4.32\times 10^{-9}}{16\times10^{-6}} = 0.27\times10^{-3}=2.7\times10^{-4}\ N/C$

Answer:

B. $2.7\times 10^{-4}\ N/C$