Question

A uniform electric field exists in the region between two oppositely charged parallel plates 1.50
A uniform electric field exists in the region between two oppositely charged parallel plates 1.50 cm apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.41�10^-6 g. A) Find the magnitude of the electric field. Use 1.60�10^-19 C for the magnitude of the charge on an electron and 1.67�10?^-27 kg for the mass of a proton. N/C B)Find the speed of the proton at the moment it strikes the negatively charged plate. m/s
apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.41×10−6 .

A) Find the magnitude of the electric field.

Use 1.60×10−19 for the magnitude of the charge on an electron and 1.67×10−27 for the mass of a proton.

------ N/C

B)Find the speed of the proton at the moment it strikes the negatively charged plate.

------ m/s

Answer 1

Concepts and reason

The main concepts used to solve this problem are the electric field, force, and kinematic equation for speed.

Initially, calculate the acceleration of the object by using the equation of motion. Later calculate the electric field by using the relationship between the force and charge and finally calculate the speed of proton.

Fundamentals

The force acting on the object due to the Newton’s second law is,

$F = ma$

Here, m is the mass and a is the acceleration of the object.

The expression for the force acting between two oppositely charged plane parallel plates is,

$F = qE$

Here, q is the charge on the plates and E is the electric field between the plates.

Equate the above two expression of the forces and simplify the expression for the electric field.

$\begin{array}{c}\\ma = qE\\\\E = \frac{{ma}}{q}\\\end{array}$

The kinematic equation for the speed is,

${v^2} = {u^2} + 2ax$

Here, u is the initial speed, v is the final speed, x is the distance, and a is the acceleration.

The kinematic equation for the distance is,

$x = ut + \frac{1}{2}a{t^2}$

Here, x is the distance, u is the initial speed, t is the time, and a is the acceleration.

(A)

The kinematic equation for the distance is,

$x = ut + \frac{1}{2}a{t^2}$

Substitute 0 for u and rearrange the equation for the a.

$\begin{array}{c}\\x = \left( 0 \right)t + \frac{1}{2}a{t^2}\\\\x = \frac{1}{2}a{t^2}\\\\a = \frac{{2x}}{{{t^2}}}\\\end{array}$

Substitute 1.50 cm for x and $1.41 \times {10^{ - 6}}{\rm{ s}}$ for t.

$\begin{array}{c}\\a = \frac{{2\left( {1.50{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}}{{{{\left( {1.41 \times {{10}^{ - 6}}{\rm{ s}}} \right)}^2}}}\\\\ = 1.509 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The expression for the electric field is,

$E = \frac{{ma}}{q}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m, $1.509 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}$ for a, and $1.60 \times {10^{ - 19}}{\rm{ C}}$ for q.

$\begin{array}{c}\\E = \frac{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)\left( {1.509 \times {{10}^{10}}{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{1.60 \times {{10}^{ - 19}}{\rm{ C}}}}\\\\ = 157.5{\rm{ N/C}}\\\end{array}$

(B)

The kinematic equation for the speed is,

${v^2} = {u^2} + 2ax$

Here, u is the initial speed, v is the final speed, x is the distance, and a is the acceleration.

Initially, the proton is at rest on the surface.

$u = 0\;{\rm{m/s}}$

Substitute 0 m/s for u in the equation ${v^2} = {u^2} + 2ax$ and solve for $v$ .

$\begin{array}{c}\\{v^2} = {\left( 0 \right)^2} + 2ax\\\\v = \sqrt {2ax} \\\end{array}$

The expression for the speed of the proton is,

$v = \sqrt {2ax}$

Substitute $1.509 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}$ for a and 1.50 cm for x.

$\begin{array}{c}\\v = \sqrt {2\left( {1.509 \times {{10}^{10}}{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {\left( {1.50{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)} \right)} \\\\ = 2.13 \times {10^4}{\rm{ m/s}}\\\end{array}$

Ans: Part A

The magnitude of the electric field is 157.5 N/C.