Question

Two point charges q1=3uC and q2=-3uC are located 20 cm apart in vacuum?​

Answer 1

Question :-

Two point charges q1=3uC and q2=-3uC are located 20 cm apart in vacuum?

Solution :-

Topic :-

Electrostatics

Given :-

Two point charges $\mathtt{q_1} = 3 μC and \mathtt{q_2} = –3 μC are located 20 cm apart in a vacuum.</p><p></p><p><strong>To Find :-</strong></p><p></p><ul><li>What is the electric field at the midpoint O of the line AB joining the two charges ?</li></ul><p></p><ul><li>If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge ?</li></ul><p></p><p><strong>Concept</strong><strong> </strong><strong>:</strong><strong>-</strong></p><p></p><p>Electric Field Intensity of a point Charge</p><p></p><p>[tex]Electric \: Field \: Intensity \: of \: a \: point \: Charge</p><p></p><p>[tex]\mathtt{\overrightarrow{\tt E}=\dfrac{\overrightarrow{\tt F}}{q_0}=\dfrac{kqq_0}{r^2\cdot q_0}\hat{r}=\dfrac{kq}{r^2}\hat{r}}[\tex]</p><p></p><p>$

Where

$</p><p>\mathtt{\overrightarrow{\tt E}=Electric \: \;Field\;Intensity}E=Electric \: Field \: Intensity</p><p></p><p>\mathtt{\overrightarrow{\tt F}=Electrostatic\;Force}F=Electrostatic \: Force</p><p></p><p>\mathtt{k=Coulomb's\;Constant=9\times 10^9\:Nm^2C^{-2}}k=Coulomb′s \: Constant=9×109Nm2C−2</p><p></p><p>\mathtt{q=Charge}q=Charge</p><p></p><p>\mathtt{q_0=Test \: \;Charge}q0=TestCharge</p><p></p><p>\mathtt{r=Distance\;of\;separation}r=Distance \: of \: separation</p><p></p><p>$

Electric Force on a Charged particle due to external Electric Field.

$</p><p></p><p>\mathtt{\overrightarrow{\tt F}=q_0\overrightarrow{\tt E}}F=q0E</p><p></p><p></p><p></p><p></p><p></p><p>$

Direction of Electric Field from a Charge.

• Direction of Electric Field from a positive charge is radially outward.

• Direction of Electric Field from a negative charge is radially inward.

Direction of Electric Force with respect to external Electric Field.

• On positive charge, Electric Force is in direction of external Electric Field.

• On negative charge, Electric Force is in opposite direction of external Electric Field.

Solution :-

(a) Distance between two charges is 20 cm. So, midpoint O will be at

$\frac{20 \: cm}{2} = 10 \: cm$

$</p><p></p><p>\mathtt{Hence,r=10\:cm=\dfrac{10}{100}\:m=0.1\:m}Hence,r=10cm=10010m=0.1m</p><p></p><p>Direction \: of \: Electric \: Field \: by \: charge  \: \mathtt{q_1}q1  \: will \: be \: radially \: outward \: as \: it \: is \: a \: positive \: charge. \: (Here, \: it \: will \: be \: in \: direction \: of \: OB)</p><p></p><p>Direction \: of \: Electric \: Field \: by \: charge \:  \mathtt{q_2}q2 will \: \: be radially \: inward \: as \: it \: is \: a \: negative \: charge. (Here, \: it \: will \: be \: in \: direction \: of \: OB)</p><p></p><p> \:$

Now

$Net \: Electric \: Field \: Intensity \: at \: point \: O = \: Magnitude \: of \: Electric \: Field \: at \: point \: O \: by \: charge \:  \mathtt{q_1}q1 with direction + Magnitude \: of \: Electric \: Field \: at \: point \: O \: by \: charge  \: \mathtt{q_2}q2  \: with \: direction</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=\overrightarrow{\tt E}_{q_1}+\overrightarrow{\tt E}_{q_2}}Enet=Eq1+Eq2</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{kq_1}{r^2}\widehat{OB}+\dfrac{kq_2}{r^2}\widehat{OB}}Enet=r2kq1OB+r2kq2OB</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{k(3\mu C)}{(0.1\:m)^2}\widehat{OB}+\dfrac{k(3\mu C)}{(0.1\:m)^2}\widehat{OB}}Enet=(0.1m)2k(3μC)OB+(0.1m)2k(3μC)OB</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{2k(3\mu C)}{(0.1\:m)^2}\widehat{OB}}Enet=(0.1m)22k(3μC)OB</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{2(9\times 10^9)(3\times 10^{-6})}{(0.1\times 0.1)}\widehat{OB}}Enet=(0.1×0.1)2(9×109)(3×10−6)OB</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=54\times10^5\;\widehat{OB}}Enet=54×105OB</p><p></p><p>\mathtt{\overrightarrow{\tt E}_{net}=5.4\times10^6\;NC^{-1}\;\widehat{OB}}Enet=5.4×106NC−1OB</p><p></p><p>$

(b) On negative charge, Electric Force is in opposite direction of external Electric Field. So, Force will be in opposite direction of OB that is in direction of OA.

The point O is same point, so we can use same magnitude of Electric Field as calculated above in formula of Electric Force on a Charged particle due to Electric Field.

So the solution is :-

$</p><p></p><p>\mathtt{\overrightarrow{\tt F}=q_0\overrightarrow{\tt E}}F=q0E</p><p></p><p>\mathtt{\overrightarrow{\tt F}=(1.5\times 10^{-9})(5.4\times 10^{6})\:\widehat{OA}\;N}F=(1.5×10−9)(5.4×106)OAN</p><p></p><p>\mathtt{\overrightarrow{\tt F}=8.1\times 10^{-3}\;\widehat{OA}\;N}F=8.1×10−3OAN</p><p></p><p>$

Answer :-

$(a) \mathtt{\overrightarrow{\tt E}_{net}=5.4\times10^6\;NC^{-1}\;\widehat{OB}}Enet=5.4×106NC−1OB</p><p></p><p>(b) \mathtt{\overrightarrow{\tt F}=8.1\times 10^{-3}\;\widehat{OA}\;N}F=8.1×10−3OAN</p><p></p><p>$

Hope you understand :)