A proton is released from rest in a uniform electric field of magnitude 8.0× 104 V/m, directed along positive x-axis. The proton undergoes a displacement of 0.30 m in the direction of the field. What is the change in electric potential of the proton as a result of this displacement?

Question

A proton is released from rest in a uniform electric field of magnitude 8.0× 104 V/m, directed along positive x-axis. The proton undergoes a displacement of 0.30 m in the direction of the field. What is the change in electric potential of the proton as a result of this displacement? (A)  2.4× 104 V  (B)  4.8× 104 V  (C)  -2.4× 104 V  (D)  -1.2× 104 V

Tardigrade Admin · Accepted Answer

Electric field of magnitude 8.0 × 104 V / m Displacement = 0.30 m Therefore, change in electric potential =-E d =-(8 × 104)(0.30) =-2.4 × 104 V

$2.4 \times 1 0^{4} V$

$4.8 \times 1 0^{4} V$

$- 2.4 \times 1 0^{4} V$

$- 1.2 \times 1 0^{4} V$

Electric field of magnitude $8.0 \times 1 0^{4} V / m$
Displacement $=$ $0.30 m$
Therefore, change in electric potential $= - E d$
$= - (8 \times 1 0^{4}) (0.30)$
$= - 2.4 \times 1 0^{4} V$

Q. A proton is released from rest in a uniform electric field of magnitude 8.0×104V/m, directed along positive x-axis. The proton undergoes a displacement of 0.30m in the direction of the field. What is the change in electric potential of the proton as a result of this displacement?

2.4×104V

4.8×104V

−2.4×104V

−1.2×104V

Electric field of magnitude 8.0×104V/m Displacement = 0.30m Therefore, change in electric potential =−Ed =−(8×104)(0.30) =−2.4×104V

$2.4 \times 1 0^{4} V$

$4.8 \times 1 0^{4} V$

$- 2.4 \times 1 0^{4} V$

$- 1.2 \times 1 0^{4} V$

Electric field of magnitude $8.0 \times 1 0^{4} V / m$
Displacement $=$ $0.30 m$
Therefore, change in electric potential $= - E d$
$= - (8 \times 1 0^{4}) (0.30)$
$= - 2.4 \times 1 0^{4} V$