Question

what is the strength of the electric field 0.020 m from a 12 μc charge? (use ( k=8.99 \times 10^9 \frac{n m^2}{c^2} ))
5.4 × 10⁶ n/c
2.7 × 10⁸ n/c
5.4 × 10¹² n/c
2.7 × 10¹⁴ n/c

Explanation:

Step1: Recall the electric field formula

The formula for the electric field \( E \) due to a point charge \( q \) at a distance \( r \) is \( E = k\frac{q}{r^2} \), where \( k = 8.99\times 10^{9}\frac{N\cdot m^{2}}{C^{2}} \), \( q \) is the charge, and \( r \) is the distance from the charge.

Step2: Convert the charge to coulombs

The charge \( q = 12\ \mu C \). Since \( 1\ \mu C=1\times 10^{- 6}\ C \), then \( q = 12\times 10^{-6}\ C=1.2\times 10^{-5}\ C \). The distance \( r = 0.020\ m \).

Step3: Substitute the values into the formula

Substitute \( k = 8.99\times 10^{9}\frac{N\cdot m^{2}}{C^{2}} \), \( q = 1.2\times 10^{-5}\ C \) and \( r = 0.020\ m \) into the formula \( E=k\frac{q}{r^{2}} \).

First, calculate \( r^{2}=(0.020\ m)^{2}=4\times 10^{-4}\ m^{2} \).

Then, \( E=(8.99\times 10^{9})\times\frac{1.2\times 10^{-5}}{4\times 10^{-4}} \).

Simplify the fraction: \( \frac{1.2\times 10^{-5}}{4\times 10^{-4}}=\frac{1.2}{4}\times 10^{-5 + 4}=0.3\times 10^{-1}=3\times 10^{-2} \).

Then, \( E = 8.99\times 10^{9}\times3\times 10^{-2} \).

Using the rule of exponents \( a^{m}\times a^{n}=a^{m + n} \), we have \( 8.99\times3\times 10^{9-2}\approx26.97\times 10^{7}=2.697\times 10^{8}\approx2.7\times 10^{8}\ N/C \).

Answer:

\( 2.7\times 10^{8}\ N/C \) (corresponding to the option "2.7 × 10⁸ N/C")