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UNO [17]
3 years ago
8

Why are force fields necessary to describe eletric force

Physics
2 answers:
Mazyrski [523]3 years ago
8 0

Answer:

electric force fields indicate the strength of the electric force that will be experienced and also the direction of the force

Explanation:

bazaltina [42]3 years ago
4 0

Answer:

Electric force is a phenomenon which is experienced by a point charge due to its presence in an electric field. Electric force fields indicate the strength of the electric force that will be experienced, and also the direction of the force. The strength is shown by the closeness of the electric field lines and the direction is shown by the direction of the lines. Therefore, force fields are necessary in describing electric force.

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Increasing the amount of current that flows through a wire ______ the strength of an electromagnet
Anestetic [448]

Electromagnet is in form of solenoid

and the magnetic field due to solenoid is given as

B = \mu_0 n i

here

i = current in the loop

so when we increase the current in electromagnet the magnetic field of the solenoid will increase

this will increase the strength of the electromagnet

so the answer would be

<em>INCREASE</em>

7 0
3 years ago
Read 2 more answers
EXERCISE 1
denpristay [2]
.........The answer is A
5 0
2 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
The periodic table arranges elements by increasing _________.
Olegator [25]

Answer:

Atomic number

Explanation:

Hope it helps you in your learning process.

4 0
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Suppose your hair grows at the rate 1/31 in. per day. find the rate at which it grows in nanometers per second. because the dist
kati45 [8]
Hair grows at rate of 9.20nm/s.
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