Question

id sphere of radius 10.0cm has a tctal positivc cnarg2 of 21,4 uc uniformly distributed throughout its volume. Calculate tho magnitude of the cicctric held 0cm from the center of tne sphere. 0.00 N/C no. is 154-520ius Irins Calculate the magnitude of the electri field 8.29cm from the center of the sphere. Calculate the magnitude of the “lectric r_id ÷0.0cm from the tenter or the sphere. Subemit AnEweTrias o10 Calculata the maanitude ot the alactriz tiald 42.0cm trom the center ot the sphera Subnit Anewwr Triss 01o An insulating ,phare is ล.ถ5am in diameter and car a 5.68uc cha ฬิ unitarmly distibuted throughout its interiar w lume. Calculata thล charge ผิndo sed by concentric sphar cal surtaca with radius r _ 2.37cm 1.16× 10-6 C r receipt no. i 134-156 อ Prenouamica Calculate the charge en closed by concentric spherical surtaca with radius r-5.63cm. İl S/L-b 음 sold conducting sphere of radius 2.J7cm has a charge 7.sguc configuration. oonducting spherical shell cf inner radius 3.93cm and outer radius 5.0Bcm s concent c ith the solid sphere and has ฮ char e 3.58u Calculate the electric field at r = 1 48cm from the center ofthis charge Submit Anewwer Tries 0/10 Calculale the electric field st3,14cin frum the ceer of this tharge onfuro Calculata tha alectric tleld at 4.47cm trom tha cantar ot thir charga cont quration. 0.DO N/C receipt no, is 154-1/9/ Calculate the clectric field atr-3.53cm from the center of this chargc contiguration. Submit Aneer Tries 010

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Answer #1

Given is:-

Radius of the sphere  r 40cm

total positive charge Q = 24.4 × 10-6C

Now,

The volume charge density of the sphere is

ho = rac{Q}{V}

by plugging all the values we get

24.4 × 10-6 π(0.40)3 p=

which gives us

ρ= 9.1 × 10-5C/

Now,

We know that

Qenclosedー〉enclosed

substituting Venclosed -Tra. A 4tr2 and all the other values we get

Ein = pr 7t eq-1

this is for the electric field inside the sphere

Therefore

part -a

For r=0

plugging this value in eq-1 we get

Ein = ON / C

Part-b

For r= 8.29 cm

(9.1 × 10-5)(8.29 × 10-2) 3(8.85× 10-12) E8.29 =

which gives us

E8.29 = 2.84 × 10°N/C

Part-c

r=40cm

For this Again using Gauss's law we get

Eoutー

by plugging all the values we get

(24.4 × 10-6) 4π(40 × 10-2)2(8.85 × 10-12) E40

which gives us

E40-1.37 × 10°M/C

part-d

for r=42 cm

(24.4 × 10-6) E42 (42 × 10-2)2(8.85 × 10-12)

which gives us

E42-1.24 × 100 M/C

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