Question

Two equal and opposite charges of magnitude 0.2 × 10^ -6 are 15 cm apart, the magnitude and direction of the resultant electric intensity E at a point midway between the charge is _______


I need steps please and the answer is 6.4 × 10 ^ 5 NC ^-1

Answer 1

Solution :

➡ Given:

✏ Magnitude of two equal and opposite charges = $\rm{0.2×10^{-6}}$ C

✏ Distance between two charges = 15cm

➡ To Find:

✏ The magnitude and direction of resultant electric intensity E at a point midway between the charges.

➡ Concept:

✏ Electric field is the property of charge particle.

✏ Electric field is vector quantity.

✏Electric field is defined as force acts per unit charge.

✏ Direction of electric field for positive charge is inside and for negative charge particle is outside.

➡ Formula:

✏ Formula of electric field at distance 'r' is given by

$\star \: \underline{ \boxed{ \bold{ \rm{ \pink{E = \frac{kq}{ {r}^{2} }}}}}} \: \star$

➡ Calculation:

✏ Distance between midpoint and a particular charge particle = half of total distance = 7.5cm

$\mapsto \rm \: {E}_{net} = E_{ + q} + E_{ - q} \\ \\ \mapsto \rm \: {E}_{net} = \frac{kq}{ {r}^{2} } \: + \frac{kq}{ {r}^{2} } \: \\ \\ \mapsto \rm \: {E}_{net} = \frac{2kq}{ {r}^{2} } \: = \frac{2 \times 9 \times {10}^{9} \times0.2 \times {10}^{ - 6} }{ ({0.075})^{2} } \: \\ \\ \mapsto \rm \: {E}_{net} = 6.4 \times {10}^{5} \: \frac{N}{C} \:$

➡ Final Answer:

$\star \rm \: Magnitude \: of \: Electric \: intensity = \\\rm \orange{6.4 \times {10}^{5} \: \frac{N}{C}} \\ \\ \star \rm \: Direction \: of \: Electric \: intensity = \\\rm \purple{ + q \: charge \: to \: - q \: charge}$

Answer 2

Answer:

Given:

2 Charges of 0.2 × 10^(-6) C are placed 15 cm apart .

To find:

Magnitude and direction of the resultant electric Field intensity at a position midpoint in-between the charges.

Concept:

Both the charges will produce an electric field intensity at the midpoint. But the direction of field due to the both the charges will be same and directed towards the negative charge.

Calculation:

$\vec E_{net} = \vec E_{1} + \vec E_{2}$

$|\vec E_{net} |= \dfrac{1}{4\pi \epsilon} \dfrac{(q1)}{ {r}^{2} } +\dfrac{1}{4\pi \epsilon} \dfrac{(q2)}{ {r}^{2} }$

Since | q1 | = | q2 | , we can say :

$|\vec E_{net} |= 2 \bigg \{\dfrac{1}{4\pi \epsilon} \dfrac{(q)}{ {r}^{2} } \bigg \}$

$|\vec E_{net} |= 2 \bigg \{9 \times {10}^{9} \dfrac{(0.2 \times {10}^{ - 6} )}{ {(0.075)}^{2} } \bigg \}$

$|\vec E_{net} |= 6.4 \times {10}^{ 5} \: N {C}^{ - 1 }$

So final answer :

$\boxed{ \blue{ \huge{ \bold{ |\vec E_{net} |= 6.4 \times {10}^{ 5} \: N {C}^{ - 1 } }}}}$