Question

A point charge of 10 mC is placed at origin. The value of electric potential at point(3, 4) m is 1.8 × 10^6 V 9 × 10^7 V 9 × 10^6 V 1.8 × 10^7 V rubbish answers will be reported

Answer 1

Answer:

Charge located at the origin, q = 8 mC= 8 × 10 - 3 C

Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = - 2 × 10 - 9 C

Point P is at a distance, d1 = 3 cm, from the origin along z-axis.

Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.

Potential at point P,    

Potential at point Q, 

Work done (W) by the electrostatic force is independent of the path.

Therefore, work done during the process is 1.27 J.

Explanation:

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Answer 2

Given :-

• Let  ABCD is a square of side √2 m. Four Electric Charges are placed at the corners of the square. Charge +10μC   is  placed at A and

• 5μC at B, -3μ C, and 8μC are placed at D.

To be found :-

• The Potential at the center of the square.

Theory :-

• The electric potential at a point define is the external work done by a unit positive charge from infinity to that point without any acceleration.

• Electric potential due to two-point - :

$\purple{\boxed{V=\frac{k\ Q}{r}}}$

Solution :-

Let the center of the square be O.

Let $\sf Q_1=10\mu\ C$

$\sf Q_2=5\mu\ C$

$\sf Q_3=-3\mu\ C$

$\sf Q_4=8\mu\ C$

From the Diagram :-

In ∆BDC

CB²+CD²=BD² [ Pythagoras theorem ]

⇒ BD²= (√2)² + (√2)²

⇒ BD² = 4m

⇒ BD = √4m

⇒ BD = 2m

Hence , OD = OB = AO = OC = 1m

We have to find the potential in the center of the square.

$\sf V{o} =V_{1} + V_{2} + V_{3} + V_{4}$

$\sf V{o}= \frac{k\ Q_1}{OA}+ \frac{k\ Q_2}{OB}+ \frac{k\ Q_3}{OC}+ \frac{k\ Q_4}{OD}$

Now put the given values!

$\sf V{o} = \frac{k \times 10 \times 10^{-6}}{1}+ \frac{k \times 5 \times 10^{-6}}{1}+\frac{k \times -3 \times 10^{-6}}{1}+\frac{k \times 8 \times 10^{-6}}{1}$

$\sf V_{o} = k \times 10^{-6}(10 + 5 - 3 + 5)$

$\sf V_{o} = k \times(15+5) \times 10^{-9}$

If the medium between the two points is air then,

$\sf\:k=\frac{1}{4\pi\:E_{o}}=9\times10^{9}Nm^{2}C^{-2}$

$\sf V_{o} = 9 \times 10^9 \times 20 \times 10^{-9}$

$\sf V_{o} = 18 \times 10^4 V$

$\sf V_{o} = 1.8 \times 10^5 V$

Therefore, the potential at the center of the square is

$\sf = 1.8 \times 10^5 V$

Hence,

• $\mathbb {Correct \;option\; 3) \;1.8 \;x \;10^5 V ✔}$

Extra Information:-

• 1) SI unit of electric potential is J/c

• 2) Electric potential is a scalar quantity