Question

An antiproton (same properties as a proton except that $q=-e$) is moving in the combined electric and magnetic fields of FIGURE. What are the magnitude and direction of the antiproton’s acceleration at this instant?

Short answer

Acceleration,$a=2.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$upward

Step-by-step solution

Magnetic attraction (part a)

Due to the electrical field $E$, the electrical force $F_E$is exerted on the charge, and it's given by

$F_E=qE$

Because antiproton charge is negative, the electrical field exerts an electrical force during a downward direction.

Equation of magnetic attraction,

$F_B=qvB$

Magnetic attraction (part b)

The magnetism is exerted upward within the wrong way of the electrical field force, In keeping with the right hand rule. As a result, the antiproton is subjected to the web force of the $2$forces.

$F_{\text{net}}=F_B-F_E$

$ma$$=qvB-qE$

$a=\frac{q(vB-E)}{m}$

Finalization

Put the values in equation

$a=\frac{q(vB-E)}{m}$

$=\frac{1.6\times {10}^{-19}\mathrm{C}\left(\right(500\mathrm{m}/\mathrm{s}\left)\right(2.5\mathrm{T})-1000\mathrm{V}/\mathrm{m})}{1.67\times {10}^{-27}\mathrm{kg}}$

$=2.4\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2$

The antiproton goes within the direction of the magnetic pull because the acceleration is positive. As a result, the acceleration is upward.