A 2 - gram object, located in a region of uniform electric field E =(300 N C -1) hat i carries a charge Q . The object released from rest at x=0, has a kinetic energy of 0.12 J at x =0.5 m. Then Q is

Question

A 2 - gram object, located in a region of uniform electric field E =(300 N C -1) hat i carries a charge Q . The object released from rest at x=0, has a kinetic energy of 0.12 J at x =0.5 m. Then Q is (A) 400 μ c (B) -400 μ c (C) 800 μ c (D) -800 μ c

Tardigrade Admin · Accepted Answer

Work done ( W )= Change in K E Force × displacement = change in KE QE × x =0.12 J Q (300) ×(0.5)=0.12 Q =(0.12/300 × 0.5)=800 μ C The charge Q is moving in the direction of electric field, hence it is positive.

Q. A $2$ - gram object, located in a region of uniform electric field $E = (300 N C^{- 1}) \hat{i}$ carries a charge $Q .$ The object released from rest at $x = 0$ , has a kinetic energy of $0.12 J$ at $x = 0.5$ $m$ . Then $Q$ is

$400 μ c$

$- 400 μ c$

$800 μ c$

$- 800 μ c$

Work done $(W) =$ Change in $K E$
Force $\times$ displacement $=$ change in $K E$
$QE \times x = 0.12 J$
$Q (300) \times (0.5) = 0.12$
$Q = \frac{0.12}{300 \times 0.5} = 800 μ C$
The charge $Q$ is moving in the direction of electric field, hence it is positive.

Q. A 2 - gram object, located in a region of uniform electric field E=(300NC−1)i^ carries a charge Q. The object released from rest at x=0, has a kinetic energy of 0.12J at x=0.5 m. Then Q is

400μc

−400μc

800μc

−800μc