Q. 564.6( 5 Votes )

# An electric field of 20 N C^{–1} exists along the x-axis in space. Calculate the potential difference V_{B} – V_{A} where the points A and B are given by,

(a) A = (0, 0); B = (4m, 2m)

(b) A = (4m, 2m); B = (6m, 5m)

(c) A = (0, 0); B = (6m, 5m)

Do you find any relation between the answers of parts (a), (b) and (c)?

Answer :

**Given:**Magnitude of Electric field: E = 20 N C

^{–1}

E is along x-axis

**Formula used:**

As Electric field is along x-axis, potential difference will be along x-direction. Which means only x co-ordinates will be considered.

We know that,

Here dV is the change in potential : dV= V

_{B}-V

_{A}

E is the electric field along positive x axis and ds is the change in displacement.

(a)

A = (0, 0); B = (4m, 2m)

∴ V

_{B}-V

_{A}= -20×(4-0) = -80 V

(b)

A = (4m, 2m); B = (6m, 5m)

∴ V

_{B}-V

_{A}= -20×(6-4) = -40 V

(c)

A = (0, 0); B = (6m, 5m)

∴ V

_{B}-V

_{A}= -20×(6-0) = -120 V

(d)

From (a),(b) and (c), we conclude that:

Potential difference of at points A = (0, 0), B = (6m, 5m)

= Potential difference at points A = (0, 0), B = (4m, 2m)

+ Potential difference at points A = (4m, 2m), B = (6m, 5m)

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