Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

7 0

3 years ago

The electronegativity values of fluorine, oxygen, and hydrogen are compared in the table. Comparison of Electronegativity Elemen

Dvinal [7]

The hydrogen bonding in H₂O is stronger than that of HF

Explanation:

Hydrogen bonds are special dipole-dipole attraction in which electrostatic attraction is established between hydrogen atom of one molecule and the electronegative atom of a neighboring molecule.

The strength of hydrogen bonds depends on the how electronegative an atom is.
Electronegativity refers to the tendency of an atom to gain electrons.
The higher the value, the higher the tendency.
This why oxygen with a higher electronegativity will form a stronger hydrogen bond with hydrogen compared to fluorine.

Learn more:

hydrogen bond brainly.com/question/12408823

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4 0

3 years ago