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scoundrel [369]
2 years ago
14

Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half way between the two plates th

e electric field has magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half way between the plates?
a. 0
b. E/2
c. 2E
d. 4E
e. E
Physics
1 answer:
melisa1 [442]2 years ago
8 0

Answer:

E

Explanation:

From the equation describing the Electric field strength of a charge

E=F/q

We can deduce that the electric field strength is dependent on the force arcing on the magnitude of the charge.

As the force increases the the electric field increases, also the electric field strength decreases with increasing charge.

Hence the distance has no effect on the electric field strength. The magnitude remain as E

You might be interested in
The x vector component of a displacement vector has a magnitude of 146 m and points along the negative x axis. The y vector comp
larisa86 [58]

Answer:  

a) the magnitude of r is  184.62

b) the direction is 37.74° south of the negative x-axis

   

Explanation:

Given the data in the question;

as illustrated in the image blow;

To find the the magnitude of r, we will use the Pythagoras theorem

r² = y² + x²

r = √( y² + x²)

we substitute

r = √((-113)² + (-146)²)

r = √(12769 + 21316 )

r = √(34085 )

r = 184.62

Therefore, the magnitude of r is  184.62

To find its direction, we need to find ∅

from SOH CAH TOA

tan = opposite / adjacent

tan∅ = -113 / -146

tan∅ = 0.77397

∅ = tan⁻¹( 0.77397 )

∅ = 37.74°

Therefore, the direction is 37.74° south of the negative x-axis

7 0
2 years ago
A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?
ycow [4]

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

6 0
3 years ago
A 65-kg woman in an elevator is accelerating upward at a rate of 0.6 m/s2. The gravitational force is ___ N?
Nastasia [14]

Answer:

The gravitational force is 130.

Explanation:

During this problem you have to multiply the 65 and the 0.6.

5 0
3 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
2 years ago
A small but measurable current of 5.8 × 10-10 A exists in a copper wire whose diameter is 3.0 mm. The number of charge carriers
swat32

Answer:

a) The current density ,J = 2.05×10^-5

b) The drift velocity Vd= 1.51×10^-15

Explanation:

The equation for the current density and drift velocity is given by:

J = i/A = (ne)×Vd

Where i= current

A = Are

Vd = drift velocity

e = charge ,q= 1.602 ×10^-19C

n = volume

Given: i = 5.8×10^-10A

Raduis,r = 3mm= 3.0×10^-3m

n = 8.49×10^28m^3

a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]

J = (5.8×10^-10) /(2.83×10^-5)

J = 2.05 ×10^-5

b) Drift velocity, Vd = J/ (ne)

Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)

Vd = (2.05×10^-5)/(1.36 ×10^10)

Vd = 1.51× 10^-5

8 0
3 years ago
Read 2 more answers
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