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Liula [17]
1 year ago
13

A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

ween the two charges?
Physics
1 answer:
aev [14]1 year ago
4 0

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

E\text{ = }\frac{k|q|}{d^{\prime2}}

The electric field strength at the midpoint due to the charge q1 will be

\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint due to the charge q2 will be

\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint between the charges will be

\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

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Answer:

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Explanation:

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s₁ = v₁t₁

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v₁ = speed of motorcycle 1 = 6.5 m/s

t₁ = time = 10 s

Therefore,

s₁ = (6.5 m/s)(10 s)

s₁ = 65 m

Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

s₂ = s₁ + 50 m

s₂ = 65 m + 50 m

s₂ = 115 m

Now, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²

where,

Vi = initial velocity of motorcycle 2 = 0 m/s

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115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

a = 230 m/100 s²

a = 2.3 m/s²

Now, using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (2.3 m/s²)(10 s)

Vf = 23 m/s

3 0
2 years ago
What is folk remedies
mario62 [17]

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3 years ago
Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc
vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= \sqrt{((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= \sqrt{(1.8*10^11) / (2.66*10^-3) = \sqrt{(6.8*10^13) = 8.25*10^6 V/m    

 

8 0
2 years ago
What are the names of the 2 charged particles in an atom and what are their charges
Sholpan [36]

Answer:

Protons and electrons

Explanation:

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  • The physical and chemical characteristics of an atom are determined by the number of these particle composition.
  • The protons present in the center of an atom called the nucleus is positively charged.
  • The neutrons are also located in the nucleus of an atom with no charge.
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An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi
BARSIC [14]

Answer:

B= 55.6×10^(-7) Tesla

Explanation:

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I : current in the wire

B= (4π×10^(-7)×125)/(2π×4.5)

B= 55.6×10^(-7) Tesla

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