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A cone is resting on a tabletop as shown in the figure with its face horizontal. A uniform electric field of magnitude 4550 n/c points vertically upward. How much electric flux passes through the sloping side surface area of the cone?.

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onyebuchinnaji

The electric flux through the slopping surface is 2.34 Nm²/C.

Electric flux through the slopping surface

The electric flux through the slopping surface is calculated as follows;

Ф = BA

where;

  • B is the electric field
  • A is the slopping area

The area of the cone is calculated as follows;

[tex]A = \pi r(r + \sqrt{h^2 + r^2} )[/tex]

where;

  • r is the radius of the cone = 2.11 cm = 0.0211 m
  • h is height of the cone = 18.5 cm = 0.185 m

[tex]A = \pi \times 0.0211 (0.0211 + \sqrt{0.185^2 + 0.0211^2} )\\\\A = 0.0137 \ m^2[/tex]

The magnetic flux is calculated as follows;

[tex]\phi = BA\\\\\phi = 4550 \times 0.0137\\\\\phi = 62.34 \ Nm^2/c[/tex]

The complete question is below;

A cone is resting on a tabletop (height = 18.5 cm and diameter is 4.22 cm) as shown in the figure with its face horizontal. A uniform electric field of magnitude 4550 n/c points vertically upward. How much electric flux passes through the sloping side surface area of the cone?.

Learn more about magnetic flux here: https://brainly.com/question/10736183

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