Question

Question 1.21: A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10 3 N/C and points radially inward, what is the net charge on the sphere?

Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-48

Answer 1

HEY BUDDY...!!!

HERE IS UR ANSWER....

_________________________

▶️ We know Electric field intensity ( E )

=> E = q / 4 π € d^2________( 1 )

▶️ Where ,

=> E = 1.5 × 10^3 N / C

=> d = distance from centre = 20 cm = 0.2 m

=> € = permittivity due to free space ( 9 × 10^9 )

⏺️ Now by eqn 1

=> q = E ( 4 π € ) d^2

⏺️ Putting values..

=> q = 1.5 × 10^3 × 0.2 × 0.2 / 9 × 10^9

=> q = 6.67 × 10^-9 C OR 6.67 n C..

HOPE HELPED..

HAVE A NICE DAY...

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Answer 2

Given, r = 20cm = 0.2 m
Electric field , E = 1.5 × 10³ N/C
radius of conducting sphere , R = 10cm = 0.1m
we know electric field is given by $\mathbb{E}=\bf{\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}}$ , where r > R

now, 1.5 × 10³ = 9 × 10^9 × q/(0.2)²
[ we know, $\bf{\frac{1}{4\pi\epsilon_0}}=9\times10^9$ ]
so, q = 1.5 × 10³ × 0.04/(9 × 10^9 )
q = (60/9) × 10^-9
q = 6.67 × 10^-9 C
q = 6.67 nC [ 1 nanoC = 10^-9C]

but here electric field points radially inwards .so, charge is negative.
e.g., charge, q = - 6.67 nC