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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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3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
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. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
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Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

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k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

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