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3 years ago

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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the

magnitude of the electric flux througha rectangular area of 1.65 m2 in the xy-plane. N m2/C

1 answer:

zzz [600]3 years ago

8 0

Answer:

The magnitude of the electric flux is $3.53\ N-m^2/C$

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area $A= 1.65 m^2$

We need to calculate the flux

Using formula of the magnetic flux

$\phi=E\cdot A$

$\phi = EA\cos\theta$

Where,

A = area

E = electric field

Put the value into the formula

$\phi=2.35\times1.65\times\cos 25^{\circ}$

$\phi=2.35\times1.65\times0.91$

$\phi=3.53\ N-m^2/C$

Hence, The magnitude of the electric flux is $3.53\ N-m^2/C$

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Read 2 more answers

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Alexxandr [17]

Answer:

a

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b

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An image representing this lever set-up is shown on the first uploaded image

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$\sum M_f = F * cos \theta * l - W cos\theta * d = 0$

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