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Keith_Richards [23]
3 years ago
5

A disk with a radius of R is oriented with its normal unit vector at an angle Q with respect to a uniform electric field. Which

of the following would result in an increase in the electric flux through the disk? Check all that apply? A. increasing the area of the disk B. decreasing the given orientation angle of the disk C. decreasing the strength of the electric field D. increasing the given orientation angle of the disk, but not exceeding an angle of 90
Physics
1 answer:
beks73 [17]3 years ago
7 0

As we know that electric flux is given by

\phi = E.A

\phi = EAcos\theta

so in order to increase the flux we have two options

1. By increasing the area of the disc

2. by changing the orientation of disc so the area of the disc is parallel to the electric field

so correct answer will be

<em>  A. increasing the area of the disk</em>

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What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

6 0
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Answer:

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Using

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So = 8.9*10^9 * 1.6*10^-19/0.6*10^-3)²

= 3.99*10^-3N/C

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