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PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359487 When one electron is taken towards the other electron, then the electric potential energy of the system

1 Becomes zero

2 Remains unchanged

3 Increases

4 Decreases

Explanation:

Potential energy of two electron system is
$= \frac{(- e) (- e)}{4 π ε_{0} r} = \frac{e^{2}}{4 π ε_{0} r}$
As $r$ decreases, potential energy increases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359488 A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system,

1 May increase or decrease

2 Increases in any case

3 Remains the same

4 Decreases in any case.

Explanation:

Case 1: For unlike charges
When one of the charges is negative, $q_{1} q_{2} < 0,$ therefore the potential energy of the system is negative.
Now, if distance between the charges is increased, potential energy of the system will increase.
Case II : For like charges (both positive or both negative) $q_{1} q_{2} > 0,$ therefore potential energy of the system is positive.
Now, if distance between the charges is increased potential energy of the system will decrease.

KCET - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359489 If electric potential energy of given system is positive, then
supporting img

1

2 Q > 3 q

2

2 Q < 3 q

3

Q > 2 q

4

Q > 3 q

Explanation:

$\frac{2 k Q^{2}}{a} - \frac{k Q q}{a} - \frac{2 k Q q}{a} > 0$
$\Rightarrow \frac{2 k Q^{2}}{a} > \frac{3 k Q q}{a}$
$\Rightarrow 2 Q > 3 q$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359490 Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side of $0.4 m .$ The potential energy of the system is

1

18 \times 10^{- 6} J

2

9 J

3

9 \times 10^{- 6} J

4

27 J

Explanation:

$U_{S y s} = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{1}}{r_{13}})$
$= 27 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359487 When one electron is taken towards the other electron, then the electric potential energy of the system

1 Becomes zero

2 Remains unchanged

3 Increases

4 Decreases

Explanation:

Potential energy of two electron system is
$= \frac{(- e) (- e)}{4 π ε_{0} r} = \frac{e^{2}}{4 π ε_{0} r}$
As $r$ decreases, potential energy increases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359488 A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system,

1 May increase or decrease

2 Increases in any case

3 Remains the same

4 Decreases in any case.

Explanation:

Case 1: For unlike charges
When one of the charges is negative, $q_{1} q_{2} < 0,$ therefore the potential energy of the system is negative.
Now, if distance between the charges is increased, potential energy of the system will increase.
Case II : For like charges (both positive or both negative) $q_{1} q_{2} > 0,$ therefore potential energy of the system is positive.
Now, if distance between the charges is increased potential energy of the system will decrease.

KCET - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359489 If electric potential energy of given system is positive, then
supporting img

1

2 Q > 3 q

2

2 Q < 3 q

3

Q > 2 q

4

Q > 3 q

Explanation:

$\frac{2 k Q^{2}}{a} - \frac{k Q q}{a} - \frac{2 k Q q}{a} > 0$
$\Rightarrow \frac{2 k Q^{2}}{a} > \frac{3 k Q q}{a}$
$\Rightarrow 2 Q > 3 q$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359490 Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side of $0.4 m .$ The potential energy of the system is

1

18 \times 10^{- 6} J

2

9 J

3

9 \times 10^{- 6} J

4

27 J

Explanation:

$U_{S y s} = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{1}}{r_{13}})$
$= 27 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359487 When one electron is taken towards the other electron, then the electric potential energy of the system

1 Becomes zero

2 Remains unchanged

3 Increases

4 Decreases

Explanation:

Potential energy of two electron system is
$= \frac{(- e) (- e)}{4 π ε_{0} r} = \frac{e^{2}}{4 π ε_{0} r}$
As $r$ decreases, potential energy increases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359488 A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system,

1 May increase or decrease

2 Increases in any case

3 Remains the same

4 Decreases in any case.

Explanation:

Case 1: For unlike charges
When one of the charges is negative, $q_{1} q_{2} < 0,$ therefore the potential energy of the system is negative.
Now, if distance between the charges is increased, potential energy of the system will increase.
Case II : For like charges (both positive or both negative) $q_{1} q_{2} > 0,$ therefore potential energy of the system is positive.
Now, if distance between the charges is increased potential energy of the system will decrease.

KCET - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359489 If electric potential energy of given system is positive, then
supporting img

1

2 Q > 3 q

2

2 Q < 3 q

3

Q > 2 q

4

Q > 3 q

Explanation:

$\frac{2 k Q^{2}}{a} - \frac{k Q q}{a} - \frac{2 k Q q}{a} > 0$
$\Rightarrow \frac{2 k Q^{2}}{a} > \frac{3 k Q q}{a}$
$\Rightarrow 2 Q > 3 q$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359490 Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side of $0.4 m .$ The potential energy of the system is

1

18 \times 10^{- 6} J

2

9 J

3

9 \times 10^{- 6} J

4

27 J

Explanation:

$U_{S y s} = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{1}}{r_{13}})$
$= 27 J$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359487 When one electron is taken towards the other electron, then the electric potential energy of the system

1 Becomes zero

2 Remains unchanged

3 Increases

4 Decreases

Explanation:

Potential energy of two electron system is
$= \frac{(- e) (- e)}{4 π ε_{0} r} = \frac{e^{2}}{4 π ε_{0} r}$
As $r$ decreases, potential energy increases.

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359488 A system of two charges separated by a certain distance apart stores electrical potential energy. If the distance between them is increased, the potential energy of the system,

1 May increase or decrease

2 Increases in any case

3 Remains the same

4 Decreases in any case.

Explanation:

Case 1: For unlike charges
When one of the charges is negative, $q_{1} q_{2} < 0,$ therefore the potential energy of the system is negative.
Now, if distance between the charges is increased, potential energy of the system will increase.
Case II : For like charges (both positive or both negative) $q_{1} q_{2} > 0,$ therefore potential energy of the system is positive.
Now, if distance between the charges is increased potential energy of the system will decrease.

KCET - 2019

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359489 If electric potential energy of given system is positive, then
supporting img

1

2 Q > 3 q

2

2 Q < 3 q

3

Q > 2 q

4

Q > 3 q

Explanation:

$\frac{2 k Q^{2}}{a} - \frac{k Q q}{a} - \frac{2 k Q q}{a} > 0$
$\Rightarrow \frac{2 k Q^{2}}{a} > \frac{3 k Q q}{a}$
$\Rightarrow 2 Q > 3 q$

PHXII02:ELECTROSTATIC POTENTIAL AND CAPACITANCE

359490 Three charges each $20 μ C$ are placed at the corners of an equilateral triangle of side of $0.4 m .$ The potential energy of the system is

1

18 \times 10^{- 6} J

2

9 J

3

9 \times 10^{- 6} J

4

27 J

Explanation:

$U_{S y s} = \frac{1}{4 π ε_{0}} (\frac{q_{1} q_{2}}{r_{12}} + \frac{q_{2} q_{3}}{r_{23}} + \frac{q_{3} q_{1}}{r_{13}})$
$= 27 J$