A thin wire carries a uniformly distributed charge Q=3.14nC along its length. The wire is bent in the form of a semi-circular arc of radius R=1m (without disturbing the charge distribution). Find the electric field at the center of curvature of the arc (in N/C ).

Question

A thin wire carries a uniformly distributed charge Q=3.14nC along its length. The wire is bent in the form of a semi-circular arc of radius R=1m (without disturbing the charge distribution). Find the electric field at the center of curvature of the arc (in N/C ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-ntwbiem4zipz1dm9.jpg" /> E=(2 k λ /R) sin(θ /2)=(2 k Q/π R2) sin(π /2)

Q. A thin wire carries a uniformly distributed charge $Q = 3.14 n C$ along its length. The wire is bent in the form of a semi-circular arc of radius $R = 1 m$ (without disturbing the charge distribution). Find the electric field at the center of curvature of the $a rc$ (in $N / C$ ).

$E = \frac{2 kλ}{R} sin \frac{θ}{2} = \frac{2 k Q}{π R ^{2}} sin \frac{π}{2}$

Q. A thin wire carries a uniformly distributed charge Q=3.14nC along its length. The wire is bent in the form of a semi-circular arc of radius R=1m (without disturbing the charge distribution). Find the electric field at the center of curvature of the arc (in N/C ).

E=R2kλ​sin2θ​=πR22kQ​sin2π​

Q. A thin wire carries a uniformly distributed charge $Q = 3.14 n C$ along its length. The wire is bent in the form of a semi-circular arc of radius $R = 1 m$ (without disturbing the charge distribution). Find the electric field at the center of curvature of the $a rc$ (in $N / C$ ).

$E = \frac{2 kλ}{R} sin \frac{θ}{2} = \frac{2 k Q}{π R ^{2}} sin \frac{π}{2}$