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Alik [6]
3 years ago
14

Charge is uniformly distributed throughout a spherical insulating volume of radius R=4.00 cm . The charge per unit volume is −6.

5 μCm3 Find the electric field at the center of the sphere. Enter a positive number if the field points radially out, negative if the field points radially in, or zero if there is no field.
Physics
1 answer:
aleksley [76]3 years ago
8 0

Answer

-8.67× 10^6 N/C

Explanation:

The Electric Field is defined as force per unit charge.

E = Q/ 4π£r2

Qv= −6.5 μCm3

Qv = Q/ V= Q/ 4/3 πr3

Hence Q = 4/3 πr3 × Qv

Hence E = 4/3 πr3 × Qv / 4π£r2= Qvr/3£

−6.5 μ × 4/ 3×8.854 ×10^-12

-6.5 × 4 × 10^6/3 = -8.67× 10^6 N/C

Note: £ = 8.854×10^-12m/F

is the permittivity of free space

Qv is the charge per unit volume

V is volume and volume

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                                             V_y,i = V*sin(30)

- Use second equation of motion in horizontal direction:

                                          x(f) = x(0) + V*cos(30)*t

                                            75 = 0 + V*cos(30)*t

                                              t = 75 /V*cos(30)

- Use equation of motion in vertical direction:

                                     y(f) = y(0) + V_y,i*t + 0.5*g*t^2

Subs the values:

                      -2 = 0 + V*sin(30)*75/V*cos(30) - 4.905*(75/Vcos(30))^2

                           -2 = 75*tan(30) - 4.905*(5625/V^2*cos^2(30))

                           V^2 = 4.905*5625 / (2 + 75*tan(30))*cos^2(30)

                                                V^2 = 812.0633

                                                 V = 28.5 m/s

- Use the third equation of motion in the interval of the throw:

                                            V^2 = U^2 + 2*a*s_a

                                               28.5^2 = 2*a*0.7

                                                a = 580 m/s^2

         

     

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