The figure shows equipotential surfaces concentric at O. The magnitude of electric field at a distance r metres from O is

Question

The figure shows equipotential surfaces concentric at O. The magnitude of electric field at a distance r metres from O is (A) (9/r2) Vm -1 (B) (16/r2) Vm -1 (C) (2/r2) Vm -1 (D) (6/r2) Vm -1

Tardigrade Admin · Accepted Answer

Potential due to point charge q at distance r is given by v=(K q/r) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0326cf02a62a2c50ee62bb4eff5bde34-.png" /> For 60 V equipotential surface, 60=(k q/r), here r =10 cm =10 × 10-2 m 60 =(k q/10 × 10-2) ∴ k q=60 × 10 × 10-2=6 ...(i) The electric field at distance r from charge q, E=(k q/r2) Substitution from Eq. (i) gives E=(6/r2) Vm -1

Q. The figure shows equipotential surfaces concentric at $O$ . The magnitude of electric field at a distance $r$ metres from $O$ is

$\frac{9}{r ^{2}} V m^{- 1}$

$\frac{16}{r ^{2}} V m^{- 1}$

$\frac{2}{r ^{2}} V m^{- 1}$

$\frac{6}{r ^{2}} V m^{- 1}$

Q. The figure shows equipotential surfaces concentric at O. The magnitude of electric field at a distance r metres from O is

r29​Vm−1

r216​Vm−1

r22​Vm−1

r26​Vm−1

Potential due to point charge q at distance r is given by v=rKq​ For 60V equipotential surface, 60=rkq​, here r=10cm=10×10−2m 60=10×10−2kq​ ∴kq=60×10×10−2=6...(i) The electric field at distance r from charge q, E=r2kq​ Substitution from Eq. (i) gives E=r26​Vm−1

Q. The figure shows equipotential surfaces concentric at $O$ . The magnitude of electric field at a distance $r$ metres from $O$ is

$\frac{9}{r ^{2}} V m^{- 1}$

$\frac{16}{r ^{2}} V m^{- 1}$

$\frac{2}{r ^{2}} V m^{- 1}$

$\frac{6}{r ^{2}} V m^{- 1}$