12th physics | Electric field due dipole along Equatorial plane | unit 1 | day 2

Added: 2026-05-05

[00:00:00]Number one electric field electrical vertical.

[00:00:39]So okay nothing but total electric field E total distance.

[00:01:28]So also equal first let us consider dipole A and B separated distance 2 A. C is a point on equatoral plane. R is a distance from O to C.

[00:01:41]And another diagram compent E plus E minus okay EU plus sin theta E vector modus theta mod it should not be negative E plus vector cos theta E vector minus S cos theta vertical componentiz.

[00:02:17]Let us consider two componentiz opposite direction. So cancel each 6846.

[00:02:36]Okay.

[00:02:42]Okay.

[00:02:43]Electal electric field E vector total electric field E plus vector cos theta plus E vector minus cos theta and PCAP minus PCAP minus PCAP minus PUS electric field equal Electric field due to an dipole.

[00:03:46]Electric field due dipole B vector E vector plus magnitude Q R² R² B square Bong B along BC okay BC square BC square BC square hypoten square b square is equal to adjent square + opposite square adjent a square + r² so e vector = 4 epsilon bc² a² + r² okay so okay So e vector minus equal to vector.

[00:05:13]Okay.

[00:05:17]From one from one. Okay.

[00:05:22]E vectoral replace - 2 E + cos theta PCAP 2 Esscapus - E theta adjent by hypotenuse hypoten Hypotenuse adjent hypoten bc a bc bc square= a square + r² Okay. BC square A= roo<unk> of A square + R² BC square root of A square + B square to - 2 E + Q / 4 pi epsilon R A² + R² formula AI roo<unk> of A² + R² Okay.

[00:07:17]B square BC square A square + R² B square A square + R square B C B C square A square + R P E vector tote - 2 P A² + R² 3X 1 + 1x 3x2 Okay. So 2 A vector E is equal to minus P vector replace P vector 4 epsilon A² + R² okay is negated A is smaller than R in two Answer E vector tot minus P vector 4 pi epsilon RQ That's