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polet [3.4K]
3 years ago
14

An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis

. The electric field strength at the point (x,y)=(0cm,10cm) is 400 N/C .
Part A) What is the charge q? Give your answer in nC.
Part B) What is the electric field strength at the point (x, y)=(10 cm, 0 cm) ? Express your answer with the appropriate units.
Physics
1 answer:
Simora [160]3 years ago
3 0

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

b)

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

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Suppose you increase your walking speed from 5 m/s to 14 m/s in a period of 3 s. What is your acceleration
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Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.
SVEN [57.7K]

The strength of electric field E is 17 N / C.

<u />

<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

                     E = F / q

where E represents the electric field strength,

           F represents the force in newton,

           q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

                   force F = 5.0 N

Electric field strength E = force / charge

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5 0
3 years ago
A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor
Kaylis [27]
<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span>U = R*i → i =  \frac{U}{R}<span>

Solving:

</span>i = \frac{U}{R}
i = \frac{12}{5}
\boxed{i = 2.4A}

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>
5 0
3 years ago
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