Two metallic spheres of radii 1cm and 3cm are given charges of −1×10−2C and 5×10−2C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
A
2×10−2C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3×10−2C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4×10−2C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1×10−2C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B3×10−2C
So given that
Charge on A Q2−1×10−2C
Charge on B Q2=5×10−2
radius of A r1=1cm
radius of B r2=5cm
Let ′q′ charge gets transfered from A to B through the conducting wire.
Charge on A =Q1−q
Charge on B =Q2+q
And now potential of both spheres become equal k(Q1−q)r1=k(Q2+q)r2 ⇒−1−q1=5+13⇒q=−2×10−2
Now charge on bigger sphere B is given by Q=Q2+q⇒Q=5−2=3×10−2=3×10−2C