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Alexxandr [17]
3 years ago
8

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

magnitude and sign) would have to be placed on a penny that has a mass of 3g to cause it to rise into the air with an upqard acceleration of 0.19 m/s^2
Physics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

q=2.997\times 10^{-4}C

Sign-Negative

Explanation:

We are given that

Electric field =E=100NC^{-1} (Radially downward)

Acceleration=0.19 ms^{-2}(Upward)

Mass of charge=3 g=3\times 10^{-3}kg

1kg=1000g

We have to find the magnitude and sign of  charge would have to be placed on a penny .

By newton's second law

\sum F_y=ma

\sum F_y=qE-mg

Substitute the values then we get

qE-mg=ma

Substitute the values then we get

q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)

100q-29.4\times 10^{-3}=0.57\times 10^{-3}

100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}

q=\frac{29.97\times 10^{-3}}{100}

q=2.997\times 10^{-4}C

Sign of charge =Negative

Because electric force acting  in opposite direction of electric field therefore,charge on penny will be negative.

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Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
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Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

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