Ask question

Ask question

All categories

2 years ago

10

A uniform electric field has magnitude E and is directed in the negative x direction. The potential difference between point a (

at x= 0.50 m ) and point b (at x= 0.85 m ) is 300 V . Part A Which point, a or b, is at the higher potential?

1 answer:

kupik [55]2 years ago

6 0

Answer:

Point b is at the higher potential

Explanation:

Since it is given that electric field is directed in the negative x direction meaning that the potential is increasing on positive x axis so b (x = 0.85) is at higher potential.

We also know that that the direction of Electric field is always from higher potential to lower potential so b must be at higher potential.

You might be interested in

Which one of the following accurately describes the force of gravity?

Aloiza [94]

B. The gravity acceleration is in the same direction as the force of gravity, and thus towards the centre of the earth

4 0

2 years ago

Which of the following statements best explains why a book resting on a table is in equilibrium?

timofeeve [1]

I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

3 0

2 years ago

Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

Where,

$I_0$ = Initial Intensity

I = Final Intensity after pass through the polarizer

$\theta$ = Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

$\theta = cos^{-1}(\frac{1}{2})^2$

$\theta = 75.52\°$

Angle with respect to maximum is $90-75.52 = 14.48\°$

8 0

2 years ago

What is the mass of an object if a force of 2N and gives it an acceleration of 2 m/s2

tino4ka555 [31]

Answer:

1kg

Explanation:

Use the formula

F=ma

m=F/a

m=2N/2m/s^2

m=1kg

6 0

2 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago