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3 years ago

In which direction does the electric field point at a position directly north of a positive charge?

Physics

2 answers:

UNO [17]3 years ago

7 0

Answer:

South

Explanation:

navik [9.2K]3 years ago

5 0

Answer:

B. South

Explanation:

An electric field can be defined as the amount of electric force per unit charge. The direction of the electric field can be determined by the motion of a positive test charge under the electric force.

The direction of electric field is radially outward for a positive charge and radially inward for a negative charge. Thus, for the electric field points toward SOUTH at a position directly south of a positive charge.

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You place a box weighing 253.1 N on an inclined plane that makes a 39.7◦ angle with the horizontal.

Masja [62]

Answer:

Explanation:

The weight of the box is

W = 253.1N

Weight is on an incline plane

θ = 39.7°

The weight of an object is always acting downward

So, the weight makes an angle of 39.7° with the vertical component

Then, it's horizontal component is

Wx = W•Sinθ

Wx = 253.1 × Sin 39.7°

Wx = 161.67 N

The horizontal component of the weight is 161.67N

This is the force acting down the plane.

Check attachment

7 0

4 years ago

A fast moving cloud of gas,ash,and rock from an erupting volcano is an what?

Lyrx [107]

Answer:

pyroclastic cloud, or pyroclastic flow

6 0

3 years ago

What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.12 m whose potential is 230

slamgirl [31]

Answer:

(a) q=3.07 nC

(b) σ=17 nC/m²

Explanation:

Given data

Radius r=0.12m

Potential V=230 V

To find

(a) Charge q

(b) Charge density σ

Solution

For Part (a)

As we know that potential is:

$V_{potential}=k_{constat}\frac{q_{charge}}{r_{radius}} \\\\q_{charge}=\frac{V_{potential*r_{radius}}}{k_{constant}}$

Substitute the given values

$q_{charge}=\frac{(0.12m)(230V)}{9*10^{9} }\\ q_{charge}=3.07*10^{-9}C\\or\\q_{charge}=3.07nC$

For Part (b)

The charge density is given by:

σ=q/(4πr²)

Substitute the given values and value of q to find charge density

$=\frac{3.07*10^{-9}C}{4\pi (0.12m)^2}\\ =1.69*10^{-8}C/m^2\\or\\=17nC/m^2$

σ=17 nC/m²

3 0

3 years ago

What characteristic of an element determines how its atoms can bond with other<br>atoms?

wlad13 [49]

Answer:

The number of electron in the outermost shell of a particular atom determine how the reactivity of atoms to form chemical bonds with other atoms

Explanation:

This is because the number of atoms in the outer shell will determine if the atom has to gain or loose electrons for it to bond with other atoms

Hope this helps, and if it does mark as branliest answer thx

8 0

3 years ago

Read 2 more answers

In this experiment you will investigate which of the following properties of Faraday's law of electromagnetic induction? (Select

11Alexandr11 [23.1K]

Answer:

Answered

Explanation:

Part A

According to Faraday's law the induced emf in coil is equal to negative of its rate of change of magnetic flux time the number of turns in the coil.

$\epsilon = -N\frac{d\phi}{dt}$ = $-N\Delta\frac{BA}{\Delta t}$

When an emf generated by a change of magnetic flux, produced current of whose magnetic field opposes the change which produces it.

By the above equation the correct options are 1,2 and 4

Part B

Large signals of frequency of 60Hz are measured by osciloscope.

Hence the correct option is part 1.

3 0

3 years ago