Chapter 26: Q. 4 (page 737)

Figure EX26.4 is a graph of $E_{x}$ . The potential at the origin is $- 50 V$ . What is the potential at $x = 3.0 m$ ?

Short Answer

Expert verified

The required potential is $- 550 V$

Step by step solution

Given information and formula used

Given :

The potential at the origin : $- 50 V$

To find : The potential at : $x = 3.0 m$

Graph -

Theory used :

Potential difference is : $∆ V = V_{f} - V_{i} = \int E_{x} \cdot ∆ x$

Calculating the required the potential difference

Using the formula to calculate potential difference, we have :

∆ V = V_{f} - V_{i} = \int E_{x} \cdot ∆ x \Rightarrow V_{3} - V_{0} = - \int_{0}^{2} E_{x} \cdot d x - \int_{2}^{3} E \cdot d x \Rightarrow V_{3} + 50 = - 200 \times 2 - \int_{2}^{3} (600 - 200 x) d x ↤ {f r o m g r a p h} \Rightarrow V_{3} = - 400 - 50 - {[600 x - \frac{200 x ²}{2}]}_{2}^{3} = - 450 - (900 - 800) = - 450 - 100 = - 550 V

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Figure EX26.4 is a graph of $E_{x}$ . The potential at the origin is $- 50 V$ . What is the potential at $x = 3.0 m$ ?

Short Answer

Step by step solution

Given information and formula used

Calculating the required the potential difference

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Figure EX26.4 is a graph of Ex. The potential at the origin is -50V. What is the potential at x=3.0m?

Short Answer

Step by step solution

Given information and formula used

Calculating the required the potential difference

One App. One Place for Learning.

Most popular questions from this chapter

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Figure EX26.4 is a graph of $E_{x}$ . The potential at the origin is $- 50 V$ . What is the potential at $x = 3.0 m$ ?