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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>

			<p>
			$7.1.3$
			A particle with a charge $q$ and a mass $m$ passes through a region of a uniform electric field of length $d$ in time $t$. The velocity $v$ of a particle entering the field is directed along the field. Determine the electric field intensity.
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		@@ -58,11 +58,11 @@
			<h3>Solution</h3>
			<p>
			The motion of particle is accelerated by electric field, so
		−	$$d = vt-\frac{at^2}{2} \;(1)$$
		+	$$d = vt-\frac{at^2}{2} \quad(1)$$
			Applying Newton Second Law
			$$F = qE = ma$$
			then,
		−	$$a = \frac{qE}{m} \;(2)$$
		+	$$a = \frac{qE}{m} \quad(2)$$
			Putting $(2)$ into $(1)$
			$$\boxed{E = 2m\frac{d-vt}{qt^2}}$$
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			<p style="text-align: right; font-style: italic; font-size: 14;">
			BSc. Luis Daniel Fernández Quintana<br>
			Physics Department (FCNE)<br>
			Universidad de Oriente, Cuba<br>
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	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
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	<h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>

	<p>		<p>
	$7.1.3$		$7.1.3$
	A particle with a charge $q$ and a mass $m$ passes through a region of a uniform electric field of length $d$ in time $t$. The velocity $v$ of a particle entering the field is directed along the field. Determine the electric field intensity.		A particle with a charge $q$ and a mass $m$ passes through a region of a uniform electric field of length $d$ in time $t$. The velocity $v$ of a particle entering the field is directed along the field. Determine the electric field intensity.
	</p>		</p>

	@@ -58,11 +58,11 @@
	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	The motion of particle is accelerated by electric field, so		The motion of particle is accelerated by electric field, so
	$$d = vt-\frac{at^2}{2} \;(1)$$		$$d = vt-\frac{at^2}{2} \quad(1)$$
	Applying Newton Second Law		Applying Newton Second Law
	$$F = qE = ma$$		$$F = qE = ma$$
	then,		then,
	$$a = \frac{qE}{m} \;(2)$$		$$a = \frac{qE}{m} \quad(2)$$
	Putting $(2)$ into $(1)$		Putting $(2)$ into $(1)$
	$$\boxed{E = 2m\frac{d-vt}{qt^2}}$$		$$\boxed{E = 2m\frac{d-vt}{qt^2}}$$
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	BSc. Luis Daniel Fernández Quintana<br>		BSc. Luis Daniel Fernández Quintana<br>
	Physics Department (FCNE)<br>		Physics Department (FCNE)<br>
	Universidad de Oriente, Cuba<br>		Universidad de Oriente, Cuba<br>
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