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1 year ago

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A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

ween the two charges?

1 answer:

aev [14]1 year ago

4 0

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

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