1.8 Two point charges $q_{A}$ = 3 μC and $q_{B}$ = –3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 $\times 10^{- 9}$ C is placed at this point, what is the force experienced by the test charge?

61 Views | Posted 5 months ago

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

1.8 (a) The situation is represented in the following figure.

O is the midpoint of AB, hence OA = OB = 10 cm = 10 $\times 10^{- 2}$ $m^{2}$

The electric field at O caused by the charge at A is given by

$E_{1}$ = $|4 π ?_{0} \times \frac{q_{A}}{r^{2}}|$ along OB and the electric field at O caused by the charge at B is given by,

$E_{2}$ = $|\frac{1}{4 π ?_{0}} \times \frac{q_{B}}{r^{2}}|$ along OB

where $?_{0}$ = Permittivity of free space = 8.854 $\times 10^{-12}$
$C^{2} N^{- 1}$ $m^{- 2}$

Net electric field at point O, E = $E_{1} + E_{2}$ = $\frac{1}{4 π ?_{0} r^{2}}$ $\times |q_{A} + q_{B}|$

$||$ $\frac{1}{4 \times π \times 8.854 \times 10^{- 12} {\times (10 \times 10^{- 2})}^{2}}$ $\times |3 \times 10^{- 6} + 3 \times 10^{- 6}|$

=5.39 $\times 10^{6}$ N $C^{- 1}$ along OB

(b) A test charge of amount 1.5 $\times 10^{- 9}$ C is placed at the midpoint O

So the force experienced by the test charge, F = q $\times E$

= 1.5 $\times 10^{- 9} \times$ 5.39 $\times 10^{6}$
N = 8.088 $\times 10^{- 3}$
N

This force is directed along the line OA, this is because th

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