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Electric Field Due to a Dipole Along the Equatorial Line

The electric ...

Question

The electric field in a region is given by $\to E = (^3 i -^4 j)$ N/C. Find out the work done (in joule) in displacing a particle of charge 1C by 1 m along the line 4y = 3x + 9___

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Solution

For $x = 0; y = \frac{9}{4}$

And $x = 1; y = 3$

$∴ \to S =^i + (3 - \frac{9}{4})^j$

$\to S =^i + \frac{3}{4}^j$

$∣ ∣ \to S ∣ ∣ = \sqrt{(1)^{2} + {(\frac{3}{4})}^{2}} = \frac{5}{4}$

$^S = \frac{4}{5}^i + \frac{3}{5}^j W =^F^S = m (\to E \to S)$

W=0

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Similar questions

\to E = (3^i - 4^j)

k g^{- 1}

1

4 y = 3 x + 9

\to E = (3^i - 4^j) N {k g}^{- 1}

4 y = 3 x + 9

\to E = (4^i - 3^j) N/kg .

2 kg

(2 m, 1 m)

(\frac{2}{3} m, 2 m)

3 x + 4 y = 10

\to E = (4^i - 3^j) N/kg .

2 kg

(2 m, 1 m)

(\frac{2}{3} m, 2 m)

3 x + 4 y = 10

\to E = (y^i +^x) N C^{- 1}

J

1 C

{\to r}_{A} = (2^i + 2 j) m

{\to r}_{B} = (4^i +^j) m

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