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Electric Charges and Fields

267922 There is a uniform electric field of strength $10^{3} V / m$ along y-axis. A body of mass $1 g$ and charge $10^{- 6} C$ is projected into the field from origin along the positive $x$ -axis with a velocity $10 m / s$ . Its speed in m/s after 10 s is (neglect gravitation)

1 10

2

5 \sqrt{2}

3

10 \sqrt{2}

4 20

Explanation:

$v = u + a t$ where $a = \frac{E q}{m}$

Electric Charges and Fields

267923 The point charges $+ 1 C, + 1 C$ and $- 1 C$ are placed at the vertices $A, B$ and $C$ of an equilateral triangle of side $1 m$ . Then
(A) The force acting on the charge at $A$ is $9 \times 10^{9} N$
(B) The electric field strength at $A$ is $9 \times 10^{9} {NC}^{- 1}$

1 A is correct but B is wrong

2

B

iscorrect butA is wrong

3 Both

A

and

B

arewrong

4 Both

A

and

B

arecorrect

Explanation:

$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}; E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$
$F = \overset{―}{F_{1}} + \overset{―}{F_{2}}; E = \overset{―}{E_{1}} + \overset{―}{E_{2}}$

Electric Charges and Fields

267941 If the electric field between the plates of a cathode ray oscilloscope be $1.2 \times 10^{4} N / C$ , the deflection that an electron will experienceif it enters at right angles to the field with kinetic energy $2000 eV$ is (T he deflection assembly is $1.5 cm$ long.)

1

0.34 cm 2

2

3.4 cm

3

0.034 mm

4

0.34 mm

Explanation:

Deflection $y = \frac{⋐ x^{2}}{4 (K)}$ whereK iskinetic energy.

Electric Charges and Fields

267942 A electric field of $1.5 \times 10^{4} {NC}^{- 1}$ exists between two parallel plates of length $2 cm$ . An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 eV$ . The deflection that the electron experiences at the deflecting plates is

1 0

0.34 mm

2

0.57 mm

3

7.5 mm 4

4

0.75 mm

Explanation:

$y = \frac{e E α^{2}}{4 k} (K = K . E)$

Electric Charges and Fields

267922 There is a uniform electric field of strength $10^{3} V / m$ along y-axis. A body of mass $1 g$ and charge $10^{- 6} C$ is projected into the field from origin along the positive $x$ -axis with a velocity $10 m / s$ . Its speed in m/s after 10 s is (neglect gravitation)

1 10

2

5 \sqrt{2}

3

10 \sqrt{2}

4 20

Explanation:

$v = u + a t$ where $a = \frac{E q}{m}$

Electric Charges and Fields

267923 The point charges $+ 1 C, + 1 C$ and $- 1 C$ are placed at the vertices $A, B$ and $C$ of an equilateral triangle of side $1 m$ . Then
(A) The force acting on the charge at $A$ is $9 \times 10^{9} N$
(B) The electric field strength at $A$ is $9 \times 10^{9} {NC}^{- 1}$

1 A is correct but B is wrong

2

B

iscorrect butA is wrong

3 Both

A

and

B

arewrong

4 Both

A

and

B

arecorrect

Explanation:

$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}; E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$
$F = \overset{―}{F_{1}} + \overset{―}{F_{2}}; E = \overset{―}{E_{1}} + \overset{―}{E_{2}}$

Electric Charges and Fields

267941 If the electric field between the plates of a cathode ray oscilloscope be $1.2 \times 10^{4} N / C$ , the deflection that an electron will experienceif it enters at right angles to the field with kinetic energy $2000 eV$ is (T he deflection assembly is $1.5 cm$ long.)

1

0.34 cm 2

2

3.4 cm

3

0.034 mm

4

0.34 mm

Explanation:

Deflection $y = \frac{⋐ x^{2}}{4 (K)}$ whereK iskinetic energy.

Electric Charges and Fields

267942 A electric field of $1.5 \times 10^{4} {NC}^{- 1}$ exists between two parallel plates of length $2 cm$ . An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 eV$ . The deflection that the electron experiences at the deflecting plates is

1 0

0.34 mm

2

0.57 mm

3

7.5 mm 4

4

0.75 mm

Explanation:

$y = \frac{e E α^{2}}{4 k} (K = K . E)$

Electric Charges and Fields

267922 There is a uniform electric field of strength $10^{3} V / m$ along y-axis. A body of mass $1 g$ and charge $10^{- 6} C$ is projected into the field from origin along the positive $x$ -axis with a velocity $10 m / s$ . Its speed in m/s after 10 s is (neglect gravitation)

1 10

2

5 \sqrt{2}

3

10 \sqrt{2}

4 20

Explanation:

$v = u + a t$ where $a = \frac{E q}{m}$

Electric Charges and Fields

267923 The point charges $+ 1 C, + 1 C$ and $- 1 C$ are placed at the vertices $A, B$ and $C$ of an equilateral triangle of side $1 m$ . Then
(A) The force acting on the charge at $A$ is $9 \times 10^{9} N$
(B) The electric field strength at $A$ is $9 \times 10^{9} {NC}^{- 1}$

1 A is correct but B is wrong

2

B

iscorrect butA is wrong

3 Both

A

and

B

arewrong

4 Both

A

and

B

arecorrect

Explanation:

$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}; E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$
$F = \overset{―}{F_{1}} + \overset{―}{F_{2}}; E = \overset{―}{E_{1}} + \overset{―}{E_{2}}$

Electric Charges and Fields

267941 If the electric field between the plates of a cathode ray oscilloscope be $1.2 \times 10^{4} N / C$ , the deflection that an electron will experienceif it enters at right angles to the field with kinetic energy $2000 eV$ is (T he deflection assembly is $1.5 cm$ long.)

1

0.34 cm 2

2

3.4 cm

3

0.034 mm

4

0.34 mm

Explanation:

Deflection $y = \frac{⋐ x^{2}}{4 (K)}$ whereK iskinetic energy.

Electric Charges and Fields

267942 A electric field of $1.5 \times 10^{4} {NC}^{- 1}$ exists between two parallel plates of length $2 cm$ . An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 eV$ . The deflection that the electron experiences at the deflecting plates is

1 0

0.34 mm

2

0.57 mm

3

7.5 mm 4

4

0.75 mm

Explanation:

$y = \frac{e E α^{2}}{4 k} (K = K . E)$

Electric Charges and Fields

267922 There is a uniform electric field of strength $10^{3} V / m$ along y-axis. A body of mass $1 g$ and charge $10^{- 6} C$ is projected into the field from origin along the positive $x$ -axis with a velocity $10 m / s$ . Its speed in m/s after 10 s is (neglect gravitation)

1 10

2

5 \sqrt{2}

3

10 \sqrt{2}

4 20

Explanation:

$v = u + a t$ where $a = \frac{E q}{m}$

Electric Charges and Fields

267923 The point charges $+ 1 C, + 1 C$ and $- 1 C$ are placed at the vertices $A, B$ and $C$ of an equilateral triangle of side $1 m$ . Then
(A) The force acting on the charge at $A$ is $9 \times 10^{9} N$
(B) The electric field strength at $A$ is $9 \times 10^{9} {NC}^{- 1}$

1 A is correct but B is wrong

2

B

iscorrect butA is wrong

3 Both

A

and

B

arewrong

4 Both

A

and

B

arecorrect

Explanation:

$F = \frac{1}{4 π ε_{0}} \cdot \frac{q_{1} q_{2}}{r^{2}}; E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}}$
$F = \overset{―}{F_{1}} + \overset{―}{F_{2}}; E = \overset{―}{E_{1}} + \overset{―}{E_{2}}$

Electric Charges and Fields

267941 If the electric field between the plates of a cathode ray oscilloscope be $1.2 \times 10^{4} N / C$ , the deflection that an electron will experienceif it enters at right angles to the field with kinetic energy $2000 eV$ is (T he deflection assembly is $1.5 cm$ long.)

1

0.34 cm 2

2

3.4 cm

3

0.034 mm

4

0.34 mm

Explanation:

Deflection $y = \frac{⋐ x^{2}}{4 (K)}$ whereK iskinetic energy.

Electric Charges and Fields

267942 A electric field of $1.5 \times 10^{4} {NC}^{- 1}$ exists between two parallel plates of length $2 cm$ . An electron enters the region between the plates at right angles to the field with a kinetic energy of $E_{k} = 2000 eV$ . The deflection that the electron experiences at the deflecting plates is

1 0

0.34 mm

2

0.57 mm

3

7.5 mm 4

4

0.75 mm

Explanation:

$y = \frac{e E α^{2}}{4 k} (K = K . E)$

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