Question

A constant electric field of 4 N/C acts along
the x-axis What is the potential difference
between points x=-2 & x=4​

Answer 1

Answer:

• The potential difference (ΔV) is - 24 Volts

Explanation:

$\rule{300}{1.5}$

Given, Constant electric field of magnitude of 4N/C is acting along the x - axis. we need to find the potential difference between the points - 2 to + 4. here as nothing is given we will take the given points/distances to be in S.I units. From the relation of Potential difference and electric field we know that,

$\longrightarrow\sf \Delta V=-\displaystyle\int\sf E\;.\;dx$

Here,

• ΔV Denotes Potential difference.
• E Denotes Electric field.
• dx Denotes small elemental displacement.

Solving,

$\longrightarrow\sf \Delta V=-\displaystyle\int\limits^{4}_{-2}\sf 4\;.\;dx\\\\\\\\\longrightarrow\sf \Delta V=-4\;\displaystyle\int\limits^{4}_{-2}\sf dx\\\\\\\\\longrightarrow\sf \Delta V=-4\Bigg[x\Bigg]^{4}_{-2}\\\\\\\\\longrightarrow\sf \Delta V=-4\times \Bigg[4-\bigg(-2\bigg)\Bigg]\\\\\\\\\longrightarrow\sf \Delta V=-4\times \Bigg[4+2\Bigg]\\\\\\\\\longrightarrow\sf \Delta V=-4\times 6\\\\\\\\\longrightarrow\sf \Delta V=-24\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf \Delta V=-24\;V}}}}$

∴ The potential difference (ΔV) is - 24 Volts.

$\rule{300}{1.5}$

Answer 2

Answer:

Here,

E

=20

i

^

⇒E

x

=20

Potential difference ,V

B

−V

A

=−∫

A

B

Edr=−∫

4

6

E

x

dx=−20∫

4

6

dx=−20(6−4)=−40V

Explanation:

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