Question

calculate the magnitude of the electric field intensity between two plates, 15 cm apart, if the p.d between them is 4.2N

Answer 1

Answer:

To determine the magnitude of the field, divide the ... 15. Critical Thinking Suppose the top charge in Figure 21-2c is a test charge measuring ... The electric field intensity between two large ... If the potential difference in problem 18 is ... between them is 4.25X103 N/C. How far ... two parallel plates that are 0.35 cm apart.

Answer 2

Concept:

• Electric field
• Potential difference
• Relation between the intensity of electric field and the potential difference
• The electric field can be calculated from the potential if the electric potential at each point in an area of space is known.
• The electric field is represented by the negative of the gradient of the electric potential in vector calculus notation.

Given:

• The potential difference between the plates V = 4.2 V
• The distance between the two plates d = 15 cm = 0.15 m

Find:

• The intensity of the electric field between the two plates

Solution:

We know that

E = dv/dx

Electric field = 4.2/0.15

Electric field = 28 V/m

The electric field intensity is 28 V/m.

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