Question

Three small charges A (+2×10^-8), B(-5×10^-8) and P (+1×10^-8) lie along a line with distances AB= 6CM, BP= 4Cm and Ap=10 cm.

(a) Calculate the force on the charge at P due to A and B.

(b) At which point X on the line connecting A and B could there be no force on the charge P due to A and B if P were placed there?.



Please no spaming​

Answer 1

Answer:

• The Force on Charge P is 6.75 × 10⁻⁵ N
• The Point X is X = 2.32 m

Given:

1. Charge on A = 2 × 10⁻⁸ C
2. Charge on B = - 5 × 10⁻⁸ C
3. Charge on C = 1 × 10⁻⁸ C

Explanation:

$\rule{300}{1.5}$

As given the Distance of Separation are,

⇒ AB = 6 cm = 0.06 m

⇒ BP = 4 cm = 0.04 m

⇒ AP = 10 cm = 0.1 m

The Charges are kept as shown in figure:-

$\setlength{\unitlength}{1.5cm}\thicklines\begin{picture}(1,1)\put(0,0){\line(1,0){4}}\put(-0.3,-0.4){A}\put(2,-0.4){B}\put(4.15,-0.4){P}\put(-0.2,0){\circle{0.4}}\put(2.1,0){\circle{0.4}}\put(4.2,0){\circle{0.4}}\end{picture}$

Now, we can see that Force on P due to A will be repulsive as they are of same signs. and, Force on P due to B will be attractive as they are of opposite signs.

As the Forces are acting at 180° angle the Resultant force will be the Subtraction of two forces,

$\setlength{\unitlength}{1.5cm}\thicklines\begin{picture}(1,1)\put(2,2){\vector(-1,0){1}} \put(2,2){\circle{0.4}}\put(2,2){\vector(1,0){1}}\put(1.95,1.5){P}\put(3.1,1.9){ F_1 }\put(0.5,1.9){ F_2 }\end{picture}$

Now, lets find the Force F₁,

From the formula,

⇒ F₁ = K (Q_{a} Q_{p}) / r²

Substituting the values,

⇒ F₁ = K × (2 × 10⁻⁸ × 1 × 10⁻⁸) / (0.1)²

⇒ F₁ = K × (2 × 10⁻¹⁶) / (10⁻²)

⇒ F₁ = K × 2 × 10⁻¹⁶ × 10²

⇒ F₁ = 9 × 10⁹ × 2 × 10⁻¹⁶ × 10²

⇒ F₁ = 9 × 10¹¹ × 2 × 10⁻¹⁶

⇒ F₁ = 18 × 10⁽¹¹⁻¹⁶⁾

⇒ F₁ = 18 × 10⁻⁵

⇒ F₁ = 1.8 × 10⁻⁴

⇒ F₁ = 1.8 × 10⁻⁴ N

Now, Finding F₂ value,

⇒ F₂ = K (Q_{b} Q_{p}) / r²

Substituting the values,

⇒ F₂ = K × (5 × 10⁻⁸ × 1 × 10⁻⁸) / (0.04)²

⇒ F₂ = K × (5 × 10⁻¹⁶) / (4 × 10⁻²)

⇒ F₂ = K × 5 × 10⁻¹⁶ × 10² / 4

⇒ F₂ = 9 × 10⁹ × 5 × 10⁻¹⁶ × 10² / 4

⇒ F₂ = 9 × 10¹¹ × 5 × 10⁻¹⁶ / 4

⇒ F₂ = 45 × 10⁽¹¹⁻¹⁶⁾ / 4

⇒ F₂ = 45 × 10⁻⁵ / 4

⇒ F₂ = 11.25 × 10⁻⁵

⇒ F₂ = 1.125 × 10⁻⁴

⇒ F₂ = 1.125 × 10⁻⁴ N

Now, let's Find the resultant force.

⇒ F = F₁ - F₂

(Here F Denotes Net Force)

⇒ F = 1.8 × 10⁻⁴ - 1.125 × 10⁻⁴

⇒ F = 10⁻⁴ (1.8 - 1.125)

⇒ F = 10⁻⁴ × 0.675

⇒ F = 6.75 × 10⁻⁵

⇒ F = 6.75 × 10⁻⁵ N

∴ The Force on Charge P is 6.75 × 10⁻⁵ N.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Point P is between A and B and we know the distance between A and B is 6 cm.

Let the Point P be at X cm Distance from A and 6 - X cm Distance from B

$\setlength{\unitlength}{1.5cm}\thicklines\begin{picture}(1,1)\put(0,0){\line(1,0){4}}\put(-0.3,-0.4){A}\put(2,-0.4){P}\put(4.15,-0.4){B}\put(-0.2,0){\circle{0.4}}\put(2.1,0){\circle{0.4}}\put(4.2,0){\circle{0.4}}\put(0.7,0.4){X}\put(3,0.4){6\;-\;X}\end{picture}$

We Know that Net Force is Zero. the force due to A is equal to force due to B.

⇒ F₁ = F₂

⇒ K (Q_{a} Q_{p}) / r₁² = K (Q_{b} Q_{p}) / r₂²

⇒ (Q_{a} Q_{p}) / r₁² = (Q_{b} Q_{p}) / r₂²

⇒ 2 × 10⁻⁸ × 1 × 10⁻⁸/(X)² = 5 × 10⁻⁸ × 1 × 10⁻⁸/(6-X)²

⇒ 2 × 10⁻¹⁶ / X² = 5 × 10⁻¹⁶ / (6-X)²

⇒ 2 / X² = 5 / (6 - X)²

⇒ 2 × (6 - X)² = 5 × X²

⇒ 2 × {36 + X² - 12 X} = 5 X²

⇒ 2 X² - 24 X + 72 = 5 X²

⇒ 3 X² + 24 X - 72 = 0

Lets Find the Discriminant,

⇒ D = b² - 4 a c

⇒ D = (24)² - 4 × 3 × - 72

⇒ D = 576 - 12 × - 72

⇒ D = 576 + 864

⇒ D = 1440.

Now, Finding X Value by Quadratic Formula,

⇒ X = ( - b ± √{D} ) / 2 a

⇒ X = ( - 24 ± √{1440} ) / 2 × 3

⇒ X = ( - 24 ± 12√{10} ) / 6

⇒ X = 6 × (- 4 ± 2√{10} ) / 6

⇒ X = - 4 ± 2√{10}

Now,

⇒ X = - 4 + 2√{10}

(Here distance cannot be negative)

⇒ X = - 4 + 6.32

⇒ X = 2.32

⇒ X = 2.32 cm

∴ The Point X from A is  X = 2.32 m.

$\rule{300}{1.5}$

Answer 2

Answer:

The Force on Charge P is 6.75 × 10⁻⁵ N

The Point X is X = 2.32 m

Given:

Charge on A = 2 × 10⁻⁸ C

Charge on B = - 5 × 10⁻⁸ C

Charge on C = 1 × 10⁻⁸ C

Explanation:

$$\rule{300}{1.5}$$

As given the Distance of Separation are,

⇒ AB = 6 cm = 0.06 m

⇒ BP = 4 cm = 0.04 m

⇒ AP = 10 cm = 0.1 m

The Charges are kept as shown in figure:-

Now, we can see that Force on P due to A will be repulsive as they are of same signs. and, Force on P due to B will be attractive as they are of opposite signs.

As the Forces are acting at 180° angle the Resultant force will be the Subtraction of two forces,

Now, lets find the Force F₁,

From the formula,

⇒ F₁ = K (Q_{a} Q_{p}) / r²

Substituting the values,

⇒ F₁ = K × (2 × 10⁻⁸ × 1 × 10⁻⁸) / (0.1)²

⇒ F₁ = K × (2 × 10⁻¹⁶) / (10⁻²)

⇒ F₁ = K × 2 × 10⁻¹⁶ × 10²

⇒ F₁ = 9 × 10⁹ × 2 × 10⁻¹⁶ × 10²

⇒ F₁ = 9 × 10¹¹ × 2 × 10⁻¹⁶

⇒ F₁ = 18 × 10⁽¹¹⁻¹⁶⁾

⇒ F₁ = 18 × 10⁻⁵

⇒ F₁ = 1.8 × 10⁻⁴

⇒ F₁ = 1.8 × 10⁻⁴ N

Now, Finding F₂ value,

⇒ F₂ = K (Q_{b} Q_{p}) / r²

Substituting the values,

⇒ F₂ = K × (5 × 10⁻⁸ × 1 × 10⁻⁸) / (0.04)²

⇒ F₂ = K × (5 × 10⁻¹⁶) / (4 × 10⁻²)

⇒ F₂ = K × 5 × 10⁻¹⁶ × 10² / 4

⇒ F₂ = 9 × 10⁹ × 5 × 10⁻¹⁶ × 10² / 4

⇒ F₂ = 9 × 10¹¹ × 5 × 10⁻¹⁶ / 4

⇒ F₂ = 45 × 10⁽¹¹⁻¹⁶⁾ / 4

⇒ F₂ = 45 × 10⁻⁵ / 4

⇒ F₂ = 11.25 × 10⁻⁵

⇒ F₂ = 1.125 × 10⁻⁴

⇒ F₂ = 1.125 × 10⁻⁴ N

Now, let's Find the resultant force.

⇒ F = F₁ - F₂

(Here F Denotes Net Force)

⇒ F = 1.8 × 10⁻⁴ - 1.125 × 10⁻⁴

⇒ F = 10⁻⁴ (1.8 - 1.125)

⇒ F = 10⁻⁴ × 0.675

⇒ F = 6.75 × 10⁻⁵

⇒ F = 6.75 × 10⁻⁵ N

∴ The Force on Charge P is 6.75 × 10⁻⁵ N.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Point P is between A and B and we know the distance between A and B is 6 cm.

Let the Point P be at X cm Distance from A and 6 - X cm Distance from B

We Know that Net Force is Zero. the force due to A is equal to force due to B.

⇒ F₁ = F₂

⇒ K (Q_{a} Q_{p}) / r₁² = K (Q_{b} Q_{p}) / r₂²

⇒ (Q_{a} Q_{p}) / r₁² = (Q_{b} Q_{p}) / r₂²

⇒ 2 × 10⁻⁸ × 1 × 10⁻⁸/(X)² = 5 × 10⁻⁸ × 1 × 10⁻⁸/(6-X)²

⇒ 2 × 10⁻¹⁶ / X² = 5 × 10⁻¹⁶ / (6-X)²

⇒ 2 / X² = 5 / (6 - X)²

⇒ 2 × (6 - X)² = 5 × X²

⇒ 2 × {36 + X² - 12 X} = 5 X²

⇒ 2 X² - 24 X + 72 = 5 X²

⇒ 3 X² + 24 X - 72 = 0

Lets Find the Discriminant,

⇒ D = b² - 4 a c

⇒ D = (24)² - 4 × 3 × - 72

⇒ D = 576 - 12 × - 72

⇒ D = 576 + 864

⇒ D = 1440.

Now, Finding X Value by Quadratic Formula,

⇒ X = ( - b ± √{D} ) / 2 a

⇒ X = ( - 24 ± √{1440} ) / 2 × 3

⇒ X = ( - 24 ± 12√{10} ) / 6

⇒ X = 6 × (- 4 ± 2√{10} ) / 6

⇒ X = - 4 ± 2√{10}

Now,

⇒ X = - 4 + 2√{10}

(Here distance cannot be negative)

⇒ X = - 4 + 6.32

⇒ X = 2.32

⇒ X = 2.32 cm

∴ The Point X from A is X = 2.32 m.

$\rule{300}{1.5}$