Question

Two point charges q1 = 3 μC and q2 = –3 μC are located 20 cm apart in a 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 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

Answer 1

Question:

Two point charges q1 = 3 μC and q2 = –3 μC are located 20 cm apart in a 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 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

Data given:

q1=3 μC

q2=–3 μC

Distance between two charges A and B=20cm

Solution:

Let O be the midpoint of A and B.

So, AO=OB=10cm

a)

Let the electric field caused by A(+3C charge) be E1 and the electric field caused by B(-3C charge) be E2. And let the net electric field be E=E1+E2.

Calculate E1 and E2. (Convert values to SI units)

$Electric field=\frac{Kq}{4 \pi \epsilon_0 ~r^2} N/c\\\\ E_1=\frac{3 \times 10^{-6}}{4 \pi \epsilon_0 ~(10 \times 10^{-2})^2} N/c (along~OB) \\\\ E_2=\frac{3 \times 10^{-6}}{4 \pi \epsilon_0 ~(10 \times 10^{-2})^2} N/c (along OA) \\\\ Net~electric~field(E) = E_1+E_2 \\\\ E= \bigg(\frac{3 \times 10^{-6}}{4 \pi \epsilon_0 ~(10 \times 10^{-2})^2} N/c+\frac{3 \times 10^{-6}}{4 \pi \epsilon_0 ~(10 \times 10^{-2})^2} N/c \bigg) \\\\ 2 \times \bigg(\frac{3 \times 10^{-6}}{4 \pi \epsilon_0 ~(10 \times 10^{-2})^2} \bigg) N/c \\\\ Replace~the~value~of~Coulomb's Constant~ \frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} \\\\2 \times \bigg[ 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(10 \times 10^{-2})^2} \bigg] N/c \\\\ 2 \times \bigg[ 9 \times 10^9 \times \frac{3 \times 10^{-6}}{10^{-2}} \bigg] N/c \\\\ 2 \times \bigg[ 9 \times 10^9 \times \frac{3 \times 10^{-\cancel{6}~4}}{\cancel{10^{-2}}} \bigg] N/c \\\\ 2 \times \bigg[ 9 \times 10^9 \times 3 \times 10^{-4}\bigg] N/c \\\\ 2 \times \[27 \times 10^5 \] =54 \times 10^5 \to \boxed{5.4 \times 10^6}$

Hence, the electric field at point O is $5.4 \times 10^6(along~OB)$

b)

Given that a test charge of magnitude $1.5 \times 10^{-9}$ is placed at O, let the force experienced by this charge be F.

We know that $F= q E$

$F= 1.5 \times 10^{-9} \times 5.4 \times 10^6~ N \\\\ \to 8.1 \times 10^3~N(along~OA)$

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Answer 2

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.

To Find :-

• What is the electric field at the midpoint O of the line AB joining the two charges ?

• 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 ?

Concept Used :-

Electric Field Intensity of a point Charge

$\mathtt{\overrightarrow{\tt E}=\dfrac{\overrightarrow{\tt F}}{q_0}=\dfrac{kqq_0}{r^2\cdot q_0}\hat{r}=\dfrac{kq}{r^2}\hat{r}}$

where

$\mathtt{\overrightarrow{\tt E}=Electric\;Field\;Intensity}$

$\mathtt{\overrightarrow{\tt F}=Electrostatic\;Force}$

$\mathtt{k=Coulomb's\;Constant=9\times 10^9\:Nm^2C^{-2}}$

$\mathtt{q=Charge}$

$\mathtt{q_0=Test\;Charge}$

$\mathtt{r=Distance\;of\;separation}$

Electric Force on a Charged particle due to external Electric Field

$\mathtt{\overrightarrow{\tt F}=q_0\overrightarrow{\tt E}}$

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 20 cm / 2 = 10 cm.

$\mathtt{Hence,r=10\:cm=\dfrac{10}{100}\:m=0.1\:m}$

Direction of Electric Field by charge $\mathtt{q_1}$ will be radially outward as it is a positive charge. (Here, it will be in direction of OB)

Direction of Electric Field by charge $\mathtt{q_2}$ will be radially inward as it is a negative charge. (Here, it will be in direction of OB)

Now,

Net Electric Field Intensity at point O = Magnitude of Electric Field at point O by charge $\mathtt{q_1}$ with direction + Magnitude of Electric Field at point O by charge $\mathtt{q_2}$ with direction

$\mathtt{\overrightarrow{\tt E}_{net}=\overrightarrow{\tt E}_{q_1}+\overrightarrow{\tt E}_{q_2}}$

$\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{kq_1}{r^2}\widehat{OB}+\dfrac{kq_2}{r^2}\widehat{OB}}$

$\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{k(3\mu C)}{(0.1\:m)^2}\widehat{OB}+\dfrac{k(3\mu C)}{(0.1\:m)^2}\widehat{OB}}$

$\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{2k(3\mu C)}{(0.1\:m)^2}\widehat{OB}}$

$\mathtt{\overrightarrow{\tt E}_{net}=\dfrac{2(9\times 10^9)(3\times 10^{-6})}{(0.1\times 0.1)}\widehat{OB}}$

$\mathtt{\overrightarrow{\tt E}_{net}=54\times10^5\;\widehat{OB}}$

$\mathtt{\overrightarrow{\tt E}_{net}=5.4\times10^6\;NC^{-1}\;\widehat{OB}}$

(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,

$\mathtt{\overrightarrow{\tt F}=q_0\overrightarrow{\tt E}}$

$\mathtt{\overrightarrow{\tt F}=(1.5\times 10^{-9})(5.4\times 10^{6})\:\widehat{OA}\;N}$

$\mathtt{\overrightarrow{\tt F}=8.1\times 10^{-3}\;\widehat{OA}\;N}$

Answer :-

So, answers are :-

(a) $\mathtt{\overrightarrow{\tt E}_{net}=5.4\times10^6\;NC^{-1}\;\widehat{OB}}$

(b) $\mathtt{\overrightarrow{\tt F}=8.1\times 10^{-3}\;\widehat{OA}\;N}$

Note : Refer to attachment for better understanding.